When 48 , L of dry N_2 gas is passed through | vedclass

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(C) The decrease in the weight of water is due to the water molecules being carried away by the $N_2$ gas. This water vapor occupies the same volume a

Vedclass · Accepted Answer

$0.034$

Vedclass · Answer

(C) The decrease in the weight of water is due to the water molecules being carried away by the $N_2$ gas. This water vapor occupies the same volume as the $N_2$ gas,which is $48 \, L$.Using the ideal gas equation $PV = \frac{m}{M}RT$,where $V = 48 \, L$,$m = 1.2 \, g$,$M = 18 \, g/mol$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 27 + 273 = 300 \, K$.$P = \frac{mRT}{MV} = \frac{1.2 	imes 0.0821 	imes 300}{18 	imes 48}$$P = \frac{29.556}{864} = 0.034 \, atm$.

When $48 \, L$ of dry $N_2$ gas is passed through $36 \, g$ of $H_2O$ at $27 \, ^oC$,the weight of water decreases by $1.20 \, g$. What is the vapor pressure of water in $atm$?

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