The volume occupied by 1.8 g of water vapour at 374°C and 1 bar pressure will be
$[Use R = 0.083 {barLK}^{- 1} {mol}^{- 1}]$

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5.37 L

96.66 L

55.87 L

3.10 L

answer is A.

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Detailed Solution

$\begin{array}{l} Above 373 K water exists as vapour and behaves like ideal gas PV = nRT \\ V = \frac{W}{M} (\frac{RT}{P}) \\ V = \frac{1.8}{18} \times \frac{0.083 \times 647}{1} = 5.37 L \end{array}$

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