Question

QUESTION 2.19

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 k Pa at 298 K. Calculate (i) molar mass of the solute (ii) vapour pressure of water at 298 K.

Answer 1

Step I : Calculation of molar mass of the solute
Case I Number of mole of solute
$\left(n_B\right)=\frac{Mass}{Molarmass}=\frac{\left(30g\right)}{\left(Mgmo{l}^{-1}\right)}=\frac{30}{M}mol$
Number of moles of water
$\left(n_A\right)=\frac{Mass}{Molarmass}=\frac{\left(90g\right)}{\left(18gmo{l}^{-1}\right)}=5mol$
Mole fraction of water
$\left(x_A\right)=\frac{n_A}{n_A+n_B}=\frac{\left(5mol\right)}{(5mol+\frac{30}{M}mol)}=\frac{M}{(6+M)}$
Vapour pressure of 1st solution $\left(p_A\right)=2.8kPa$
According to Raoult's law, $p_A=p_A^{\circ }{x}_A$
$\left(2.8kPa\right)=p_A^{\circ }\times \frac{M}{(6+M)}\dots \dots \left(i\right)$
Case II Number of moles of solute $\left(n_B\right)=\frac{30}{M}mol$
Number of moles of water
$\left(n_A\right)=\frac{Mass}{Molarmass}=\frac{\left(108g\right)}{\left(18gmo{l}^{-1}\right)}=6mol$
Mole fraction of water $\left(x_A\right)=\frac{n_A}{n_A+n_B}=\frac{\left(6mol\right)}{(6mol+\frac{30}{M}mol)}=\frac{M}{(5+M)}$
Vapour pressure of solution $\left(p_A\right)=2.9kPa$
According to Raoult's law, $p_A=p_A^{\circ }{x}_A$
$\left(2.9kPa\right)=p_A^{\circ }\times \frac{M}{5+M}\dots \dots \left(ii\right)$
Dividing Eq. (i) by Eq. (ii)
$\frac{\left(2.8kPa\right)}{\left(2.9kPa\right)}=\frac{(5+M)}{(6+M)}(0.9655\times 6)+0.9655M=5+M0.0345M=0.793M=\frac{0.793}{0.0345}=23gmo{l}^{-1}$
Step II: Calculation of vapour pressure of water
According to Rauolt's law, $p_A=p_A^{\circ }{x}_A$
or $\left(2.8kPa\right)=p_A^{\circ }=\frac{M}{(6+M)}\dots \dots \left(iii\right)$
Placing the value of M in Eq. (i)
$2.8kPa=p_A^{\circ }\frac{\left(23gmo{l}^{-1}\right)}{(6+23)gmo{l}^{-1}}{p}_A^{\circ }=\frac{2.8\times 29}{23}$
$=\frac{81.2}{23}=3.53kPa$