MHT CET · Chemistry · Solutions
\(38 \cdot 4 \mathrm{~g}\) of unknown substance (molar mass \(384 \mathrm{~g} \mathrm{~mol}^{-1}\) ) and \(116 \mathrm{~g}\) of acetone is used to prepare a solution at \(313 \mathrm{~K}\). If vapour pressure of pure acetone (molar mass \(58 \mathrm{~g} \mathrm{~mol}^{-1}\) ) is \(0.842\) atmosphere, what is the vapour pressure of solution?
- A \(0.650 \mathrm{~atm}\)
- B \(0.880 \mathrm{~atm} .\)
- C \(0.7999 \mathrm{~atm} .\)
- D \(0.958 \mathrm{~atm}\)
Answer & Solution
Correct Answer
(C) \(0.7999 \mathrm{~atm} .\)
Step-by-step Solution
Detailed explanation
\(\mathrm{n}_{2}=\frac{38.4}{384}=0.1 \mathrm{~mol}\) solute
\(\mathrm{n}_{1}=\frac{116}{58}=2 \mathrm{~mol}\) solvent (acetone)
Mole fraction \(; x_{1}=\frac{n_{1}}{n_{1}+n_{2}}=\frac{2}{2+0.1}=0.95\)
By Raoult's law, the vapour pressure of the solution is given by
\(P=x_{1} \times P_{0}=0.95 \times 0.842 \mathrm{~atm}=0.7999 \mathrm{~atm}\)
\(\mathrm{n}_{1}=\frac{116}{58}=2 \mathrm{~mol}\) solvent (acetone)
Mole fraction \(; x_{1}=\frac{n_{1}}{n_{1}+n_{2}}=\frac{2}{2+0.1}=0.95\)
By Raoult's law, the vapour pressure of the solution is given by
\(P=x_{1} \times P_{0}=0.95 \times 0.842 \mathrm{~atm}=0.7999 \mathrm{~atm}\)
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