GPT-6 Astra
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a) 40 g; b) 75%; c) 180 g·mol⁻¹; d) 3,0 mol·L⁻¹.
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Como pensou
**Calculating item development**
I need to work on developing a straightforward item in Portuguese that involves fractions like .6, .3, and .1 related to mass and purity values. I should clarify my assumptions about impurities and check if they contribute ions. I’ll compute the total cations and consider their dissociation. I seem to need around 250-450 words for this answer without excess definitions, focusing instead on the final consistent line that summarizes all salts involving one cation.
A quantidade total de cátions na solução deve ser:
n_{\text{total}}=CV=2\cdot0{,}500=1{,}00\ \text{mol}.
Como a proporção molar Q:R:T é 6:3:1, temos:
n_{Q^+}=\frac{6}{10}\cdot1=0{,}60\ \text{mol},
\qquad
n_{R^{2+}}=0{,}30\ \text{mol},
\qquad
n_{T^{3+}}=0{,}10\ \text{mol}.
Cada mol dos respectivos sais fornece um mol do cátion correspondente.
a) Massa utilizada do sal QX
A massa de QX puro necessária é:
m_{\text{puro}}=nM=0{,}60\cdot60=36\ \text{g}.
Considerando a pureza de 90\%:
0{,}90m=36
\quad\Rightarrow\quad
\boxed{m=40\ \text{g}}.
b) Pureza do sal RX₂
A massa de RX₂ puro necessária é:
m_{\text{puro}}=0{,}30\cdot150=45\ \text{g}.
Como foram utilizados 60\ \text{g} da amostra:
P\%=\frac{45}{60}\cdot100\%
\quad\Rightarrow\quad
\boxed{P\%=75\%}.
c) Massa molar do sal TX₃
A massa de TX₃ puro utilizada foi:
m_{\text{puro}}=20\cdot0{,}90=18\ \text{g}.
Portanto:
MM=\frac{m_{\text{puro}}}{n}
=\frac{18}{0{,}10}
\quad\Rightarrow\quad
\boxed{MM=180\ \text{g·mol}^{-1}}.
d) Concentração do ânion X⁻
Cada mol de QX, RX₂ e TX₃ fornece, respectivamente, 1, 2 e 3 mol de X^-. Assim:
n_{X^-}=0{,}60+2(0{,}30)+3(0{,}10)=1{,}50\ \text{mol}.
Logo:
[X^-]=\frac{1{,}50}{0{,}500}
=\boxed{3{,}0\ \text{mol·L}^{-1}}.
Resposta final: a) 40 g; b) 75%; c) 180 g·mol⁻¹; d) 3,0 mol·L⁻¹.