Question

Derive the mathematical expression for the relation between molar mass of solute and relative lowering of vapour pressure of dilute solution.

Answer 1

Explanation:

Vapor pressure of pure solvent = $p^o$

Vapor pressure of solution = $p_s$

Number of moles pf solvent= $n_1$

Number of moles of solute = $n_2$

Molar mass of solvent =$M_1$

Molar mass of solute =$M_2$

Mass of solvent =$w_1$

Mass of solute =$w_2$

According to Raoult's law:

$\frac{p^o-p_s}{p^o}=\frac{n_2}{n_1+n_2}$..(1)

$n_1=\frac{w_1}{M_1},n_2=\frac{w_2}{M_2}$..(2)

Substituting value from(2) in (1)

$\frac{p^o-p_s}{p^o}=\frac{\frac{w_2}{M_2}}{\frac{w_1}{M_1}+\frac{w_2}{M_2}}$

For dilute solution $n_2<<n_1$, so:

$\frac{p^o-p_s}{p^o}=\frac{\frac{w_2}{M_2}}{\frac{w_1}{M_1}}$

The final expression comes out to be:

$\frac{p^o-p_s}{p^o}=\frac{w_2\times M_1}{w_1\times M_2}$

Answer 2

Answer:

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