If the intermolecular forces vanish away, the volume occupied by the molecules contained in $4.5 \,kg$ water at standard temperature and pressure will be
A$5.6\,{m^3}$
B$4.5\,{m^3}$
C$11.2 \,litre$
D$11.2\,{m^3}$
Medium
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A$5.6\,{m^3}$
a $\mu=\frac{\text { Mass of water }}{\text { Molecular wt. of water }}=\frac{4.5 kg }{18 \times 10^{-3} kg }=250$
$T =273 K$ and $P =10^{5} N / m ^{2}( STP )$
From $\quad PV =\mu RT$
$\Rightarrow V =\frac{\mu RT }{ P }=\frac{250 \times 8.3 \times 273}{10^{5}}=5.66 \; m ^{3}$
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