Consider the following equilibrium reaction at $27 ° C$
$N a_{2} S O_{4} .10 H_{2} O (s) ⇌ N a_{2} S O_{4} (s) + 10 H_{2} O (g); K_{p} = 1.024 \times 10^{- 27} a t m^{10}$
The vapour pressure of water at $27 ° C$ is 0.04 atm.
Choose the incorrect statement from amongst the following:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$N a_{2} S O_{4} .10 H_{2} O (s)$ effloresces at a relative humidity below 5%.

$N a_{2} S O_{4} (s)$ can be used to dehumidify atmosphere having relative humidity more than 5%.

The mass of a sample of $N a_{2} S O_{4} .10 H_{2} O (s)$ kept in a large confinement having partial pressure of water $0.01 a t m,$ decreases till it attains a constant weight.

The mass of a sample of $N a_{2} S O_{4} (s)$ kept in a large confinement having partial pressure of water $0.04 a t m,$ increases till it attains a constant weight.

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$N a_{2} S O_{4} .10 H_{2} O (s) ⇌ N a_{2} S O_{4} (s) + 10 H_{2} O (g); K_{p} = 1.024 \times 10^{- 27} a t m^{10}$

$p (H_{2} O) = 0.002 a t m$

Corresponding $R H = (0.002 / 0.04) \times 100 = 5 %$
$N a_{2} S O_{4} .10 H_{2} O (s)$ effloresces below a RH of 5%
$N a_{2} S O_{4} (s)$ keeps on absorbing moisture, till RH is reduced to 5%.
$If in air, p (H_{2} O) < 0.002 a t m N a_{2} S O_{4} .10 H_{2} O (s) effloresces.$
If in air, $p (H_{2} O) > 0.002 a t m N a_{2} S O_{4} (s)$ absorbs water (acts as a dehumidifier)