Question

For the equilibrium CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(g)

Kp = 2.25 × 10–4atm2 and vapour pressure of water is 22.8 Torr at 298 K.

CuSO4.5H2O(s) is efflorescent (i.e., loses water) when relative humidity is :

• A. less than 33.3%
• B. less than 50 %
• C. less than 66.6%
• D. above 66.6%

Answer 1

Answer: (B)

less than 50 %

Answer 2

$CuSO_{4}.5H_{2}O(s)$ $\rightleftharpoons$ $CuSO_{4}.3H_{2}O(s) + 2H_{2}O(g)K_{p} = 2.25 \times 10^{4}$

$K_{P} = {P^{2}}_{H_{2}O} = 2.25 \times 10^{4}$

$P_{H_{2}O} = 1.5 \times 10^{- 2}$

Vapour Pr =$\frac{22.8}{760} = 3 \times 10^{2}$

$P.H = \frac{P_{H_{2}O}}{V.P.} \times 100 = 50\%$

Therefore, (2) option is correct.