Answer
a) $64\frac{g}{mol}$
b) $C_{2}H_{5}Cl$
Work Step by Step
$\omega (C)=37.23 \% $
$\omega (H)=7.81 \% $
$\omega (Cl)=54.96 \% $
$T=150^{\circ}C=(150+273.15)K=423.15K$
$p=1atm=101.325kPa$
$V=500ml = 0.5dm^{3}$
$m=0.922g$
---a) $M=?$
b) molecular formula?
a) $pV=nRT \implies pV=\frac{m}{M}RT \implies pMV=mRT \implies M=\frac{mRT}{pV}=\frac{0.922g\times 8.314\frac{J}{K\times mol}\times 423.15 K}{101.325 kPa \times 0.5 dm^{3}}=64\frac{g}{mol}$
b) $\omega (C)=37.23 \% $
$0.3723\times 64 \frac{g}{mol}\approx 24 \frac{g}{mol}=2\times 12\frac{g}{mol}=2\times A(C)$. Hence, this compound contains $2$ carbon atoms.
$\omega (H)=7.81 \% $
$0.0781\times 64 \frac{g}{mol}\approx 5 \frac{g}{mol}=5\times 1\frac{g}{mol}=5\times A(H)$. Hence, this compound contains $5$ hydrogen atoms.
$\omega (Cl)=54.96 \% $
$0.5496\times 64 \frac{g}{mol}\approx 35.2 \frac{g}{mol}=1\times 35.2\frac{g}{mol}=1\times A(Cl)$. Hence, this compound contains $1$ chlorine atom.
Therefore, the molecular formula of this volatile liquid is $C_{2}H_{5}Cl$.
