Explanation

(a)
Ideal gas equation is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Vp = nRT$
When $V$ and $T$ are same,
$p \propto n$
Thus, when number of moles, $i.e.,\ n$ is least, it will exert least pressure.
$(a)\ n = \frac{wt.}{mol.wt.} = \frac{0.0355}{33.5} = 1 \times 10^{- 3}\ mol$
$(b)\ n = \frac{0.071}{33.5} = 2 \times 10^{- 3}\ mol$
$(c)\ n = \frac{number\ of\ molecules\ }{N_{A}}$
$\ \ \ \ \ \ \ \ \ \ = \frac{6.023 \times 10^{21}}{6.023 \times 10^{23}} = 0.01\ mol$
$(d)\ n = 0.02\ mol$
Thus, 0.0335 g chlorine will exert the least pressure.