A) 13g
B) 26g
C) 6.5 g
D) None of these
Correct Answer: C
Solution :
At temperature 300 K, the saturation vapour pressure is 3.6kPa for unit volume, i.e. 1m3 \[pV=nRT=\frac{m}{M}RT\]where, m = mass of vapour and M = molar weight of water Thus,\[m=\frac{MpV}{RT}=\frac{18\times 3.3\times {{10}^{3}}\times 1}{8.3\times 300}\]\[\simeq 26g\] As, relative humidity is 25% Therefore, the amount of mercury present in unit volume is\[26\times 0.25=6.5\text{ }g\]You need to login to perform this action.
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