Answer
a) $p=71.6atm$
b) $\rho = 5.9 \frac{kg}{m^{3}}$
Work Step by Step
$n(H_{2})=36mol$
$V=12.3L = 12.3 dm^{3}$
$T=25^{\circ}C=(25+273.15)K=298.15K$
a) According to the ideal gas equation, $pV=nRT$
$\implies p=\frac{nRT}{V}=\frac{36mol\times 8.314\frac{J}{K\times mol}\times 298.15K}{12.3dm^{3}}=7255.08kPa\approx 7.26\times 10^{6} Pa=\frac{7.26\times 10^{6}}{101325}atm=71.6atm$
Here, we have used the fact that if we put the volume in ideal gas equation expressed in $dm^{3}$, we will obtain the pressure in $kPa$ unit.
b) $pV=n(H_{2})RT \iff pV=\frac{m(H_{2})}{M(H_{2})}RT$
If we multiply both sides by $\frac{M(H_{2})}{V}$, we will obtain:
$pV\times \frac{M(H_{2})}{V}=\frac{m(H_{2})}{M(H_{2})}RT\times \frac{M(H_{2})}{V} \iff pM(H_{2})=\frac{m(H_{2})}{V}RT \iff pM(H_{2})=\rho (H_{2})RT$
Therefore, $\rho (H_{2})=\frac{pM(H_{2})}{RT}=\frac{7.26\times 10^{6} Pa\times 2\frac{g}{mol}}{8.314\frac{J}{K\times mol}\times298.15 K}=5857.63\frac{g}{m^{3}}\approx 5.9 \frac{kg}{m^{3}}$
