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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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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%

Can you answer this question?
 
 

1 Answer

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Answer: (b) above 60.5%
 
$ 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 %

 

answered Nov 16, 2013 by mosymeow_1
edited Nov 23, 2013 by mosymeow_1
 

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