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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?

  1. A 0.89
  2. B 0.81
  3. C 0.79
  4. D 0.71
Verified Solution

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…