| \[N{{a}_{2}}HP{{O}_{4}}.12{{H}_{2}}O(s)N{{a}_{2}}HO{{P}_{4}}.\] \[7{{H}_{2}}O(s)+5{{H}_{2}}O(g)\] |
| Given: |
| (i) \[{{K}_{p}}=3.2\times {{10}^{-14}}at{{m}^{5}}\] |
| (ii) The vapour pressure of water at \[0{}^\circ C=6.0\times {{10}^{-3}}atm\] |
| I. The hydrated salt will act as efflorescent when exposed to the air at \[0{}^\circ C\] below 33% relative humidity. |
| II. The dehydrated salt will act as deliquescent when exposed to the air at \[0{}^\circ C\] above 33% relative humidity. |
| III. The dehydrated salt will act as efflorescent when exposed to the air at \[0{}^\circ C\] below 33% relative humidity. |
| IV. The hydrated salt will act as deliquescent when exposed to the air at \[0{}^\circ C\] above 33% relative humidity. |
A) I, II
B) III, IV
C) I, III
D) II, IV
Correct Answer: A
Solution :
[a] \[Kp=p{{({{H}_{2}}O)}^{5}}=3.2\times {{10}^{-14}}=32\times {{10}^{-15}}\] \[\therefore \,\,\,\,P({{H}_{2}}O)=5\sqrt{32\times {{10}^{-15}}}\] \[=2.0\times {{10}^{-3}}atm.\] Relative humidity \[=\left( \frac{2.0\times {{10}^{-3}}}{6.0\times {{10}^{-3}}} \right)\times 100=33.3%\] Therefore the hydrated salt \[(N{{a}_{2}}.HP{{O}_{4}}.12{{H}_{2}}O)\] will lose water (i,e, efflorescent) when exposed to the air at \[0{}^\circ C\] below \[33.3%\] relative humidity. The dehydrated salt \[(N{{a}_{2}}HP{{O}_{4}}.7{{H}_{2}}O)\] will absorb moisture (i.e., deliquescent) when exposed to the air at \[0{}^\circ C\] above \[33.3%\]relative humidity.
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