Question

(a) Explain the following
(i) Henry's law about a dissolution of a gas in a liquid.
(ii) Boiling paint elevation constant for a solvent.
(b) A solution of glycerol $\left(C_3{H}_8{O}_3\right)$ in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of ${100.42}^{o}C$. What mass of glycerol was dissolved to make this solution? ($K_b$ for water =0.512 K kg mol${}^{-1}$)

Answer 1

Henry's law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If $p$ is the partial pressure of the gas in the vapour phasur phase and $x$ is the mole fraction of the gas, then Henry's law can expressed as:

$p=K_{H}x$

where,

$K_H$ is Heny's law constant

(ii) The boiling point elevation constant or molal elevation constant is a constant quality for a solute which is related to molar mass and elevation in boiling point by the following relation

$K_b=\frac{\Delta T_b\times M_B\times W_A}{W_B\times 1000}$

Where, $K_b$ is the boiling point elevation constant

$M_B$ is the molar mass of the solute

$W_B$ is the weight of the solute

$W_A$ is the weight of the solvent

$\Delta T_b$ is the elevation in boiling point

(b) $W_B=?$

$W_A=500g$

$K_b=0.512Kkgmo{l}^{-1}$

$\Delta T_b=100.42{\hat{A}}^{o}C-100{\hat{A}}^{o}C$

$=0.42{\hat{A}}^{o}C$

$M_B=3\times 12+8\times 1+3\times 16$

$=36+8+48=92$

$\Delta T_b=\frac{K_b\times W_b\times 1000}{M_B\times W_A}$

$0.42=\frac{0.512\times W_B\times 1000}{92\times 500}$

$\frac{0.42\times 92\times 500}{0.512\times 1000}=W_B$

$W_B=37.33g$