TS EAMCET · Chemistry · Solutions
At T(K), 0.1 moles of a non-volatile solute was dissolved in 0.9 moles of a volatile solvent. The vapour pressure of pure solvent is 0.9 bar. What is the vapour pressure (in bar) of solution?
- A 0.89
- B 0.81
- C 0.79
- D 0.71
Answer & Solution
Correct Answer
(B) 0.81
Step-by-step Solution
Detailed explanation
\begin{array}{l}{r}{\mathrm{n}_{\mathrm{A}}=0.1 mole } \\\mathrm{n}_{\mathrm{B}}=0.9 \mathrm{~mole} \\\chi_{\mathrm{A}}=\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{A}}+\mathrm{n}_{\mathrm{B}}}=\frac{0.1}{0.9+0.1}=0.1…
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