Questão 25

**Analyzing solubility dynamics** I need to solve for the Portuguese queries related to solubility. I'm working with oil and aqueous phases, focusing on the densities and solubility at 90°C. There’s a need to determine how much of compound X dissolves in water and whether it remains saturated as temperatures change. I'll calculate the aqueous phase density as it relates to the solubility of X, considering the observations at different temperatures and check if the conditions show saturation with undissolved X present.**Examining phase densities** At 10°C, the solubility is 80g, resulting in a phase mass of 180g, which gives a density of 1.2 g/mL. Since the oil density is 1.4, the aqueous phase floats on top. Densities change linearly with temperature. I’ll calculate where their densities equalize using equations for aqueous and oil densities. It seems they intersect around 20°C. I need to explain that the undissolved X indicates saturation and consider whether the solute amount is sufficient under these conditions.

I'm working out the oil density as a linear function of temperature, giving ρ_oil(T) = 1.5 − 0.01T, and now setting up the aqueous phase with 150 mL volume, 100 g water, and X dissolved at its saturated solubility. Computing the aqueous density at both endpoints gives ρ_aq(T) = 1.1 + 0.01T. Setting ρ_oil = ρ_aq yields T = 20 °C, and checking the endpoints confirms the phases swap positions there—oil floats above at high temperature, sinks below at low temperature—so the inversion temperature is 20 °C, answer A.

Setting up the aqueous phase density as a function of temperature using the saturation data at 90°C and 10°C, then doing the same for the oil phase. Setting the two linear density expressions equal to each other gives x=10, so the temperature where densities match is 20°C.

I'm computing the aqueous phase density as a function of temperature using the solubility data at 10°C and 90°C, getting a linear relation of 1.2 + 0.01(T-10) g/mL, and I'm setting up the analogous linear expression for the oil phase density, which decreases from 1.4 at 10°C to 0.6 at 90°C. Setting the two density expressions equal gives T=20°C, which matches option A, and checking the phase ordering (oil on top at 90°C, aqueous on top at 10°C) confirms this is consistent with the given figure.

The problem is in Portuguese about a hypothetical experiment. The task is to find the temperature where the phase order inverts. At 90°C, oil sits above water plus X. At 10°C, water plus X sits above oil. Both phases have linear density changes with temperature. The aqueous volume and mass remain constant. The given data includes oil density at 0.6 g/mL and 1.4 g/mL, plus X solubility of 200 g/100 g H₂O at 90°C and 80 g/100 g H₂O at 10°C. The inversion occurs when the two phase densities are equal. The aqueous phase density depends on the temperature-dependent mass of dissolved X, which changes because the solution remains saturated with undissolved X. The aqueous phase is saturated at every temperature because undissolved X remains. The solubility data at 90°C and 10°C must be used to determine the aqueous density as a function of temperature. The aqueous density can be calculated directly at 90°C and 10°C using the saturated mass of X. Because the problem states that both phase densities behave linearly with temperature, the aqueous density will also vary linearly between those two points. The inversion temperature is simply where this linear aqueous density curve intersects the linear oil density curve. The aqueous density at 90°C is 2 g/mL and at 10°C is 1.2 g/mL. The oil density at 90°C is 0.6 g/mL and at 10°C is 1.4 g/mL. Both phase densities vary linearly with temperature. The inversion occurs when these two linear curves intersect. The linear equations for both phase densities are now fully derived. The oil density is 1.4 minus 0.01 times (t minus 10), and the aqueous density is 1.2 plus 0.01 times (t minus 10). Setting them equal yields t equals 20°C. The answer is A. The inversion occurs at 20°C. The problem is multiple choice with options A through E. This seems too simple for an ITA exam. Perhaps the solubility data is not linear. The linear density assumption is explicitly stated, so the solubility data does not need to be linear between the given temperatures. The numbers are chosen so that the two linear curves intersect exactly at 20°C. This confirms the answer is A.

We need answer question. Need solve. Let's restate. Need determine temperature at which inversion of order of phases occurs. System initially at 90°C: oil (top), water + X (bottom). Upon cooling to 10°C, diagram shows water + X top, oil bottom. So densities: at high T, oil less dense than aqueous phase; at low T, oil more dense than aqueous phase? Actually inversion of order means which phase on top/bottom. Initially oil top, aqueous bottom => density oil < density aqueous at 90°C. Finally oil bottom, aqueous top => density oil > density aqueous at 10°C. So crossover temperature where densities equal? Need find temperature where density of oil equals density of aqueous phase (water + X). Densities vary linearly with T. Oil density linear: at 90°C 0.6 g/mL, at 10°C 1.4 g/mL. Aqueous phase density? Contains water + dissolved X? Also non-solubilized X? Need account solubility changes, mass of water constant 100 g, volume aqueous phase constant 150 mL? Wait statement: "Suponha que o volume da fase aquosa (água + X) permaneça constante e igual a 150 mL, que a massa de água nessa fase seja de 100g e que tal massa também permaneça constante." So aqueous phase volume = 150 mL constant; water mass = 100 g constant. It contains water + X (dissolved? and maybe undissolved? "X não solubilizado" at bottom indicates excess solid X not in solution? Let's parse. Diagram shows initially oil top, water + X bottom, and arrow "X não solubilizado" pointing from bottom layer? Actually text: "X não solubilizado" under both diagrams maybe indicating undissolved X? Hmm image: there is a black bottom layer labeled? Let's inspect: Left: top oil, bottom "água + X"; below bottom layer an arrow labeled "X não solubilizado". Right: top "água + X"; bottom oil; below oil arrow "X não solubilizado". So there is undissolved X (solid?) at bottom? Maybe "X não solubilizado" means undissolved X in bottom phase? Actually arrow from bottom points down, label "X não solubilizado" maybe indicates solid X not solubilized present at bottom? But if phase inversion, undissolved X maybe stays with aqueous phase? Hmm. Need determine temperature at which density of oil equals density of aqueous phase. Aqueous phase consists of water + dissolved X (maybe saturated solution) and possibly undissolved X? But "X não solubilizado" might be excess solid X (not dissolved) at bottom, not part of aqueous phase? The aqueous phase (água + X) may be saturated solution with undissolved solid? The diagram labels bottom layer "água + X" and arrow "X não solubilizado" maybe separate solid? Let's read: "Esse sistema é resfriado de 90 °C a 10 °C. Suponha que o volume da fase aquosa (água + X) permaneça constante e igual a 150 mL, que a massa de água nessa fase seja de 100g e que tal massa também permaneça constante. Considere que as densidades das duas fases apresentam um comportamento linear nesse intervalo de temperatura. Assinale a alternativa que corresponde à temperatura em que ocorre a inversão de ordem das fases presentes no sistema." So two phases: oil phase (pure oil? density given) and aqueous phase (water + X). The mass of water in aqueous phase 100g constant. Volume of aqueous phase 150 mL constant. Solubility of X at 90°C = 200 g/100g H2O; at 10°C = 80 g/100g H2O. Since cooling, solubility decreases, so some X precipitates (não solubilizado). But "X não solubilizado" may be solid X in bottom? It may not be part of aqueous phase? However density of aqueous phase? The aqueous phase includes dissolved X and water, but excludes precipitated solid? The volume of aqueous phase constant 150 mL, mass of water constant 100g. If dissolved X amount changes with T, density of aqueous phase changes due to dissolved X mass. But if undissolved X is present as separate solid, the aqueous phase is saturated solution; density of saturated solution depends on concentration (solubility) and temperature. The density of aqueous phase can be calculated from total mass of solution (water + dissolved X) divided by volume (150 mL). The dissolved X mass equals solubility at that temperature (if saturated) because excess solid present. At 90°C solubility 200g/100g water => dissolved X = 200g. At 10°C solubility 80g/100g water => dissolved X = 80g. So mass aqueous phase = 100 + dissolved X. Volume constant 150 mL? But if dissolved X changes, volume might change, but problem says volume aqueous phase remains constant 150 mL? It says "Suponha que o volume da fase aquosa (água + X) permaneça constante e igual a 150 mL, que a massa de água nessa fase seja de 100g e que tal massa também permaneça constante." So volume of aqueous phase (including dissolved X) constant 150 mL; water mass constant 100g. Thus density aqueous = (100 + dissolved X)/150. At 90: (100+200)/150 = 300/150 = 2 g/mL. At 10: (100+80)/150 = 180/150 = 1.2 g/mL. Oil density at 90 = 0.6 g/mL, at 10 = 1.4 g/mL. At 90: oil density 0.6 < aqueous 2 => oil top (correct). At 10: oil density 1.4 > aqueous 1.2 => oil bottom (correct). Density equal when oil density = aqueous density. Oil density linear: ρ_oil(T) = a + b T? Let's compute slope: from 10°C to 90°C, ΔT=80°C, Δρ = 0.8 g/mL (1.4-0.6=0.8), slope = 0.8/80 = 0.01 g/mL per °C. So ρ_oil(T) = 0.6 + 0.01*(T-90)? Actually at T=90, 0.6; at T=10, 0.6 +0.01*(10-90)=0.6-0.8=-0.2? Wait 0.01*( -80)= -0.8, sum -0.2? Wrong. Let's set slope positive? As T increases, density decreases? Data: at 90°C 0.6, at 10°C 1.4. So density decreases with increasing T: slope = (0.6-1.4)/(90-10)= -0.8/80 = -0.01 g/mL/°C. Equation: ρ_oil(T) = 1.4 + (-0.01)*(T-10) = 1.4 -0.01(T-10) = 1.5 -0.01 T? At T=10: 1.4; at T=90: 1.5 -0.9 =0.6 yes. So ρ_oil(T)=1.5 -0.01 T (°C). Aqueous density? It changes with dissolved X, which depends on solubility S(T) = g X per 100 g water. Solubility likely linear? Problem only gives two points: at 90°C 200 g/100g, at 10°C 80 g/100g. It says densities of two phases present linear behavior in interval? Actually "Considere que as densidades das duas fases apresentam um comportamento linear nesse intervalo de temperatura." The densities, not solubility. The aqueous phase density depends on dissolved X concentration, which may not be linear? But maybe they assume density of aqueous phase linear with temperature? Or solubility linear? Need infer. They give solubility at two temperatures; maybe solubility also linear? The problem asks temperature of inversion; need solve. We need decide aqueous phase density at intermediate T. If solubility is linear? Let's examine. Aqueous phase density = mass/volume = (100 g water + dissolved X mass)/150 mL. Dissolved X mass = solubility S(T) if saturated. Since solid X present, solution saturated at all T. So density = (100 + S(T))/150. S(T) maybe can be assumed linear? They didn't state solubility linear, only densities of two phases linear? But "densidades das duas fases apresentam um comportamento linear" might mean density of each phase vs temperature is linear. For oil, density linear. For aqueous phase, density also linear? If solubility not linear, density might not be linear. But maybe because mass dissolved changes linearly? Let's test with data: S at 90=200, at 10=80. If linear S(T) = 80 + (200-80)*(T-10)/80 = 80 + 1.5*(T-10) = 1.5T +65? At T=10: 80; T=90: 200. Then aqueous density = (100 + S)/150 = (100 + 1.5T +65)/150 = (165 +1.5T)/150 = 1.1 +0.01 T? At 10: 1.1+0.1=1.2; at 90:1.1+0.9=2.0. Linear. Then equality with oil: 1.5 -0.01T = 1.1 +0.01T => 0.4 =0.02T => T=20°C. That corresponds option A 20°C. That seems plausible. But let's verify if solubility linear is intended? The problem states "Considere que as densidades das duas fases apresentam um comportamento linear nesse intervalo de temperatura." It does not explicitly say solubility linear. But density of aqueous phase linear if dissolved mass linear? Maybe they expect using average? Let's see if other interpretations yield other options. We need be careful. Alternative: aqueous phase density maybe constant? No, because dissolved X changes. But maybe "X não solubilizado" is not part of aqueous phase? They may consider aqueous phase only water + dissolved X, volume constant 150 mL, mass water 100g. Dissolved X at each T equals solubility? Yes if excess solid present. The density at T might be computed by interpolation of solubility? Since they gave two points, likely assume linear variation of solubility (or density) with T. The phrase "densidades das duas fases apresentam comportamento linear" maybe means oil density and aqueous phase density both linear, so we can use linear interpolation. To find aqueous density at arbitrary T, we can interpolate between densities at 90 and 10? We have only two points for aqueous density: at 90 = 2 g/mL, at 10 = 1.2 g/mL. If we assume aqueous density also linear with T, then same equation: ρ_aq(T) = 1.2 + (2.0-1.2)*(T-10)/80 = 1.2 +0.01*(T-10)=1.1 +0.01T? Wait at T=90: 1.2 +0.01*80=2.0 yes. So same. Thus equality: oil density linear between 0.6 and 1.4: ρ_oil(T)=1.4 + (0.6-1.4)*(T-10)/80 = 1.4 -0.01*(T-10)=1.5 -0.01T. Equality: 1.5 -0.01T = 1.1 +0.01T => 0.4 =0.02T => T=20°C. So answer A. But need ensure no other factors: Maybe volume of aqueous phase constant 150 mL, but as X precipitates, volume occupied by solid? However problem says volume of aqueous phase (água + X) remains constant 150 mL; undissolved X maybe separate? Actually "X não solubilizado" maybe not in aqueous phase? But if undissolved solid is present, the volume of aqueous phase (solution) remains 150 mL, while solid volume maybe excluded? The wording "volume da fase aquosa (água + X)" suggests the aqueous phase includes water + dissolved X, not solid X? But "X não solubilizado" maybe solid at bottom separate? The diagram labels bottom "água + X" and "X não solubilizado" arrow? Let's examine image: Left: top layer gray "óleo"; bottom layer white "água + X"; below bottom layer a downward arrow with text "X não solubilizado". That might mean "X não solubilizado" is the solid X that settles at bottom, not part of aqueous phase? Actually if it's not solubilized, it may be solid X at the bottom, below the aqueous phase? But diagram shows "água + X" as one layer, and "X não solubilizado" maybe indicates that the X in that layer is not solubilized? Hmm. Let's read original: "Esse sistema é resfriado de 90 °C a 10 °C. Suponha que o volume da fase aquosa (água + X) permaneça constante e igual a 150 mL, que a massa de água nessa fase seja de 100g e que tal massa também permaneça constante." So phase aqueous contains water + X. They don't mention undissolved X as separate phase; perhaps "X não solubilizado" is just label for solid X in equilibrium? Actually "X não solubilizado" could be the portion of X that did not dissolve, which may be present as solid in the bottom of the aqueous phase? But "fase aquosa (água + X)" maybe includes both dissolved and undissolved X? If undissolved X is suspended? Hmm. Let's consider if undissolved X is not part of aqueous phase but a separate solid phase, then density of aqueous phase should be based only on dissolved X. That's what we did. At 90°C solubility 200g/100g, so 200g dissolved; any extra X? The problem doesn't give total X amount, only solubility. The diagram "X não solubilizado" maybe indicates excess solid X present at all temperatures? Actually if initial system had excess X? They label "X não solubilizado" at bottom, so undissolved X exists. But the amount doesn't matter for saturated solution density. However if undissolved solid X is part of the aqueous phase volume? The volume of aqueous phase (water + X) might include undissolved X? But "X não solubilizado" is separate? Let's parse Portuguese: "X não solubilizado" = X not solubilized. It may be the solid that precipitates and is not in solution. The diagram: left: top oil, bottom "água + X"; arrow below "X não solubilizado" maybe points to solid X at bottom? Actually the arrow is below the bottom layer, labeled "X não solubilizado" with a curved arrow? In image, there's a small black rectangle at bottom with an arrow pointing down/right, label "X não solubilizado". It might represent a solid phase (undissolved X) at the bottom. On right, after cooling, oil at bottom, aqueous top; again "X não solubilizado" arrow below oil? Wait right diagram: top "água + X"; bottom "óleo"; below oil an arrow "X não solubilizado". That would mean solid X is below oil? That seems odd; if oil is denser, it goes to bottom, but undissolved X maybe even denser? Actually "X não solubilizado" might be solid X that settles at the bottom, below oil? Let's inspect: On left, oil top, aqueous bottom, and "X não solubilizado" arrow below aqueous? On right, water+X top, oil bottom, and "X não solubilizado" arrow below oil? That suggests solid X is always at very bottom, below both liquid phases. So undissolved X is a separate solid phase at bottom, not part of aqueous phase. The aqueous phase is the solution (water + dissolved X) above the solid. Its volume constant 150 mL, water mass 100g. The dissolved X amount changes with T. So our calculation holds. But wait, if undissolved X is separate solid, its volume? The problem says "volume da fase aquosa (água + X) permaneça constante e igual a 150 mL" so the solution volume is 150 mL, not including solid. Good. Now, need solve carefully: At 90°C, solubility 200g/100g water, so dissolved X = 200g; aqueous phase mass = 300g; volume 150 mL => density 2 g/mL. At 10°C, solubility 80g/100g => mass = 180g; volume 150 mL => density 1.2 g/mL. Oil density at 90 = 0.6, at 10 = 1.4. Densities linear. Find T where ρ_oil = ρ_aq. Let's derive with linear equations. Let T in °C. Oil density: ρ_o(T) = 0.6 + (1.4-0.6)*(90 - T)/(90-10)? Actually as T increases from 10 to 90, density decreases from 1.4 to 0.6. So ρ_o(T) = 1.4 + (0.6-1.4)*(T-10)/(90-10) = 1.4 -0.8*(T-10)/80 = 1.4 -0.01(T-10) = 1.5 -0.01T. Aqueous density: At 90: 2.0; at 10: 1.2. Linear: ρ_aq(T) = 1.2 + (2.0-1.2)*(T-10)/80 = 1.2 +0.8*(T-10)/80 = 1.2 +0.01(T-10) = 1.1 +0.01T. Set equal: 1.5 -0.01T = 1.1 +0.01T => 0.4 =0.02T => T=20°C. Yes. But let's double-check if solubility linear? If not, but problem states densities linear, so aqueous density linear with T. We used endpoints. Good. However, potential twist: The aqueous phase density may include undissolved X? If "água + X" includes all X (dissolved + undissolved) in the phase? But they separately label "X não solubilizado" maybe not part of aqueous phase? If undissolved X is included in aqueous phase volume, then density would be different. Let's examine: Suppose aqueous phase volume 150 mL includes water + all X (dissolved + undissolved) present in that phase? But "X não solubilizado" may be in the aqueous phase? If all X (including undissolved) is in the aqueous phase (suspension), then mass of aqueous phase = water + total X (dissolved + undissolved). But we don't know total X amount; only solubility. If excess solid X present, total X in phase could be more than solubility. But problem says "X não solubilizado" maybe not part of aqueous phase? The diagram labels "X não solubilizado" separately, so likely separate solid phase. So our assumption is correct. Let's consider another interpretation: At 90°C, solubility 200g/100g, so 200g dissolved. At 10°C, solubility 80g/100g, so 80g dissolved; 120g precipitated as solid. The solid X may settle at bottom, separate. The aqueous phase volume constant 150 mL, water mass 100g. So density of aqueous phase at 10 = (100+80)/150 = 1.2. Yes. Potential issue: Volume of aqueous phase may change due to dissolved X? But problem explicitly says volume constant 150 mL. So density just mass/volume. Let's ensure oil density interpolation: They say "densidades das duas fases apresentam comportamento linear nesse intervalo de temperatura." Could mean both densities as function of temperature are linear. So yes. Let's see if any alternative answer from using solubility linear but not density linear? Actually aqueous density linear if solubility linear; but if solubility not linear, density may not be linear; but problem states densities linear, so okay. But what if they expect using average density? Let's test if using mass of X at 90 and 10 but volume constant, then density at T could be computed via linear interpolation of density, which is equivalent to linear solubility? Wait if solubility not linear but density linear, we don't need solubility linear. We already used density linear. Let's verify endpoints with volume 150 mL: At 90: water 100g + dissolved X 200g = 300g /150 mL = 2 g/mL. At 10: 100+80=180g /150=1.2 g/mL. Yes. Now, the inversion temperature: At 20°C, ρ_oil = 1.5 -0.01*20 = 1.3 g/mL; ρ_aq = 1.1 +0.01*20 = 1.3 g/mL. At T<20, oil density > aqueous? Let's check T=10: oil 1.4 > aqueous 1.2 => oil bottom. At T>20, oil density < aqueous? T=90: oil 0.6 < aqueous 2.0 => oil top. So inversion at 20°C. Option A. Need produce final answer in Portuguese with explanation. Include line "Resposta: A". The user says "Resolva a questão e explique o seu raciocínio em português. Termine com uma linha neste formato: Resposta: X". So final should include explanation and answer. Let's craft explanation step by step. Potential nuance: The question says "inversão de ordem das fases presentes no sistema." Could the inversion occur when densities equal, but also due to volume? Yes. Let's include calculations: - Densidade do óleo: linear. Use points (90,0.6) and (10,1.4). Equation: ρ_óleo(T)=1.5 -0.01T. - Fase aquosa: massa de água 100g; massa de X dissolvido = solubilidade em g/100g água. At 90: 200g; at 10: 80g. Volume constante 150 mL. Densidade em 90: (100+200)/150 = 2.0 g/mL; em 10: (100+80)/150 = 1.2 g/mL. Linear: ρ_aq(T)=1.1 +0.01T. - Equal: 1.5 -0.01T = 1.1 +0.01T => T=20°C. Check if at 20°C, dissolved X =? Solubility linear? At 20°C, S = 80 + (200-80)*(20-10)/80 = 80 +120*10/80 = 80+15=95 g? Then density = (100+95)/150=195/150=1.3 g/mL. Yes. Let's also mention that as T decreases below 20, oil becomes denser than aqueous phase, so oil goes to bottom; above 20, oil less dense, top. Thus inversion at 20°C. Need ensure units: Temperatures in °C. Let's formulate final answer elegantly. Potential issue: "densidades das duas fases apresentam comportamento linear" Could the oil density linear but aqueous density maybe constant? No, because dissolved X changes. But if they only state densities linear, our interpolation is valid. Let's present. Let's write in Portuguese. Final line: Resposta: A

**Analyzing the Scenario** I'm currently focused on dissecting the hypothetical experiment presented. My initial assessment centers on the observed phase separation after temperature change. I'm noting the initial layer configuration of oil above water plus 'X', and the inversion after cooling, with 'X' now deposited below. This change in stratification strongly suggests a shift in the relative densities and solubilities. I'm exploring the implications of the temperature change on the components and looking for the defining characteristics of 'X'. **Calculating Phase Inversion** I'm now diving into the quantitative aspects. I've noted the constant aqueous phase volume and water mass. I'm focusing on the solubilities of 'X' at different temperatures, the linear density changes of the oil, and the implication of X always being saturated. Determining the temperature point where the density inversion happens between the two liquids is my current focus to solve the problem. **Interpreting Linear Behavior** I'm now clarifying the implications of "linear density behavior." It confirms both phase densities change linearly with temperature. I've re-examined the given data, and am focusing on how the solubility of X, combined with the constant aqueous phase volume, determines the density of that phase at different temperatures. I'm calculating the mass and subsequently, the density of the aqueous phase at 90°C and 10°C, to find the tipping point. **Determining Critical Density** I've just calculated the aqueous phase densities. At 90°C, it's 2.0 g/mL, and at 10°C, it's 1.2 g/mL, derived from X's solubility. I've also re-examined the oil's densities. Oil is 0.6 g/mL at 90°C and 1.4 g/mL at 10°C. My next step is to use the linear relationships of these densities with temperature to find the crossover point where the densities are equal. **Calculating Crossover Point** Now I am calculating the crossover point, I'm using the linear equations derived for the aqueous and oil phases. I've formulated the equations representing density as a function of temperature for both. I am currently focusing on solving the equation where these two functions intersect to determine the temperature at which the phase inversion occurs.