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Question

If $${ CuSO }_{ 4 }\cdot 5H_{ 2 }O\left( s \right) \rightleftharpoons { CuSO }_{ 4 }\cdot { 3H }_{ 2 }O\ (s)+{ 2H }_{ 2 }O\ (l){ K }_{ p }=1.086\times { 10 }^{ -4 }{ atm }^{ 2 }\ at\ { 25 }^{ o }C$$ The efflorescent nature of $${CuSO}_{4}\cdot \ {5H}_{2}O$$ can be noticed when vapour pressure of $$ {H}_{2}O$$ in atmosphere is:


A
>7.92 mm
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B
<7.92 mm
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C
7.92 mm
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D
none of these
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Solution

The correct option is B $$<7.92\ mm$$
Efflorescent nature of $$CuSO_4.5H_2O$$ means the loss of some water molecules so the compound can float on the surface rather than at bottom.

The pressure constant contains only the partial pressure of gas, the partial solid and liquid is taken unity.

So, $$CuSO_4.5H_2O(s)\leftrightharpoons CuSO_4.3H_2O(S)+2H_2O(g)$$
$$K_p=(P_{H_2O})^2$$
$$(P_{H_2O})^2=1.086\times 10^{-4}atm^2$$
$$P_{H_2O}=1.04\times 10^{-2}atm$$

Now, $$1atm=760\:mm\:of\:Hg$$
$$P_{H_2O}=7.92\:mm\:of\:Hg$$
So, for the reaction to proceed forward, the $$P_{H_2O}<7.92\:mm\:of\:Hg$$
Hence, the correct option is (B).


Chemistry

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