Question

Calculate the vapour pressure of solution having 3.42 g of canesugar in 180 g water at ${40}^{o}C$ and ${100}^{o}C$. Given that, the boiling point of water is ${100}^{o}C$ and heat of vaporisation is 10 kcal $mo{l}^{-1}$ in the given temperature range. Also, calculate the lowering in vapour pressure of 0.2 molal cane-sugar at ${40}^{o}C$:

• A. $58$ mm
• B. $13$ mm
• C. $78$ mm
• D. none of these

Answer 1

Answer: (A)

$58$ mm

At ${100}^{o}C$, Vapour pressure of pure water $\left(P^0\right)=760mm$

$\because \frac{P^0-P_S}{P_S}=\frac{w\times M}{m\times W}$ .....(i)

$\therefore \frac{760-P_S}{P_S}=\frac{3.42\times 18}{342\times 180}$

$\therefore P_S=759.2mm$

Also we have, $2.303log\frac{P_2}{P_1}=\frac{\Delta H[T_2-T_1]}{R{T}_1{T}_2}$

$\because P_2=760mm;T_2=373K$;

$T_1=313K$ and $\Delta H=10kcalmo{l}^{-1}$

$\therefore$ From eq. (ii), $2.303log\frac{760}{P_1}=\frac{10}{2\times {10}^{-3}}\times \frac{[373-313]}{373\times 313}$
$\therefore P_1=58.2mm$ at 313 K

Now at ${40}^{o}C:\frac{P^0-P_S}{P_S}=\frac{m\times W}{m\times W}$

$\frac{58.2-P_S}{P_S}=\frac{3.42\times 18}{352\times 180}$

$\therefore P_S=58.14mm$

For $0.2$ molal solution: $P_{H_{2}O}^0=58.2mm$ at ${40}^{o}C$

$\because w/m=0.2$ and $W=1000g;M=18$

$\therefore \frac{58.2-P_S}{P_S}=\frac{0.2\times 18}{1000}$

$\therefore P_S=57.99mm\approx 58mm$

Hence, option A is correct.