A) 52 g
B) 31.2 g
C) 26 g
D) 5.2 g
Correct Answer: B
Solution :
[b] \[PV=\frac{w}{m}RT\](for vapours of \[{{H}_{2}}O\]) \[P=3.6\times {{10}^{3}}Pa;v=2\times {{10}^{-3}}{{m}^{3}};T=300K\] \[\therefore w{{}_{{{H}_{2}}O}}=\frac{3.6\times {{10}^{3}}\times 18\times 2\times {{10}^{-3}}}{8.314\times 300=0.052}\] \[w{{}_{{{H}_{2}}O}}\]=52g Since, relative humidity=60%, therefore, amount of \[{{H}_{2}}O=52\times 0.6=31.2g\]
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