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# Given $NaSO_4.10H_2O (s) \rightleftharpoons NaSO_4 (s) + 10H_2O (g) ; K_p =4.08 \times 10^{-25}$ and vapour pressure of water at $0^\circ$C = 4.58 Torr. At what relative humidity will $Na_2SO_4$ be deliquescent when exposed to the air at $0^\circ$C ?

(a) below 60.5%

(b) above 60.5%

(c) above 39.5%

(d) below 39.5%

$Na_2SO_4.10H_2O(s) \rightleftharpoons Na_2SO_4(s) + 10H_2O(g)$

$K_p = 4.08 \times 10^{-25}$

$\therefore (\rho_{H_2O})^{10} = 4.08 \times 10^{-25}$
$\rho_{H_2O} = (4.08 \times 10^{-25})^{\frac{1}{10}} = x (say)$

$log x = \frac{1}{10} log(4.08 \times 10^{-25})$
$= \frac{1}{10} [log 4.08 + log 10^{-25}]$
$= \frac{1}{10}v[log 4.08 -25]$
$= \frac{1}{10} [0.0611 - 25]$
$= 0.0611 - 2.5 = \bar{3}.5611$

$x = antilog \bar{3}.5611 = 3.64 \times 10^{-3}$

$Na_2SO_4 (s)$ will absorb water if $R_H$ is more than $\frac{3.64 \times 10^{-3}}{4.58 / 760} \times 100$ $= \frac{0.364 \times 760}{4.58}$ = 60.4 %

edited Nov 23, 2013