Explanation

(b)
Given, mass of C=10.5 g
H=1.0 g
$p$=1atm
$V$= 1 L
$T = 127℃$=127+273=400 K
Mass of gas=2.81 g
Weight of C + weight of hydrogen=10.5+1.0=11.5 g
$\therefore$ % of carbon=$\frac{10.5}{11.5} \times 100 = 91.3\%$
$\therefore$ % of hydrogen=$\frac{1.0}{11.5} \times 100 = 8.7$%
[table] Element | % | At.
weight | Ratio of atoms | Simplest ratio ; C
H | 91.3
8.7 | 12
1 | 91.3/12=7.61
8.7/1=8.7 | 7.61/7.61=1
8.7/7.61
=1.14$\times$7=8 [/table]
From gas equation, $pV = nRT$
or $n = \frac{pV}{RT}$
$\frac{mass}{mole\ mass} = \frac{pV}{RT}$
or 2.81/mole mass=$\frac{1 \times 1}{0.082 \times 400}$
=92
Empirical formula wt.=$C_{7}H_{8}$
$\therefore$ Empirical formula=$7 \times 12 + 8 \times 1$
$= 92$
$n = \frac{molecular\ wt.}{empirical\ formula\ wt.} = \frac{92}{92} = 1$
Molecular formula=$n$ (empirical formula)
$= 1\ \left( C_{7}H_{8} \right)$
$= C_{7}H_{8}$