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Standard XII

Chemistry

Relative Lowering of Vapour Pressure

The vapour pr...

Question

The vapour pressure of pure benzene at $50^{o} C is 268 m m of H g$ . How many moles of non-volatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure $167 m m of H g$ at $50^{o} C$ ?

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Solution

Applying Raoult's law in the following form:
$\frac{P_{A}^{\circ} - P_{S}}{P_{S}} = \frac{w_{B} \times M_{A}}{w_{A} \times M_{B}} = \frac{w_{B} / M_{B}}{w_{A} / M_{A}}$
$w_{A} = mass of A (Solvent)$
$w_{B} = mass of B (Solute)$
$M_{A} = molar mass of A$
$M_{B} = molar mass of B$
$\Rightarrow \frac{w_{B} / M_{B}}{w_{A} / M_{A}}$ = Number of moles of solute per mole of benzene
$\Rightarrow \frac{n_{B}}{n_{A}} = \frac{P_{A}^{\circ} - P_{S}}{P_{S}} \Rightarrow \frac{n_{B}}{n_{A}} = \frac{(268 - 167)}{167} = 0.6047 \approx 0.605$

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Relative Lowering of Vapour Pressure

Standard XII Chemistry

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The vapour pressure of pure benzene at 50oC is 268 mm of Hg. How many moles of non-volatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure 167 mm of Hg at 50oC ?

Applying Raoult's law in the following form: P∘A−PSPS=wB×MAwA×MB=wB/MBwA/MA wA=mass of A (Solvent) wB=mass of B (Solute) MA=molar mass of A MB=molar mass of B ⇒wB/MBwA/MA= Number of moles of solute per mole of benzene ⇒nBnA=P∘A−PSPS⇒nBnA=(268−167)167=0.6047≈0.605

The vapour pressure of pure benzene at $50^{o} C is 268 m m of H g$ . How many moles of non-volatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure $167 m m of H g$ at $50^{o} C$ ?