For very dilute solutions relative lowering of vapour pressure is equal to

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$\frac{P^{0}}{P^{0}} = \frac{n_{1}}{n_{1} + n_{2}}$

$\frac{P^{0} - P}{P^{0}} = \frac{n_{1}}{n_{1} + n_{2}}$

$\frac{P^{0} - P}{P^{0}} = \frac{W_{1}}{M_{1}} \times \frac{M_{2}}{W_{2}}$

$\frac{P^{0} - P}{P^{0}} = \frac{n_{2}}{n_{1} + n_{2}}$

answer is A.

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Detailed Solution

According to Raoult’s law,

$\frac{P^{0} - P}{P^{0}} = \frac{n_{1}}{n_{1} + n_{2}}$

For very dilute solutions $n_{2} > > n_{1}$

$\frac{P^{0} - P}{P^{0}} = \frac{n_{1}}{n_{2}}$

$n_{1} = number of moles of solute = \frac{W_{1}}{M}$

$n_{2} = number of moles of solvent = \frac{W_{2}}{M_{2}}$

$\frac{P^{0} - P}{P^{0}} = \frac{\frac{W_{1}}{M_{1}}}{\frac{W_{2}}{M_{2}}} = \frac{W_{1}}{M_{1}} \times \frac{M_{2}}{W_{2}}$

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