Ammonia has $T_{c} = 405 K$ and $P_{c} = 111.5 atm$ . Using $b = R T_{c} /8 P_{c}$ with $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the value of $b$ in $mL mo l^{- 1}$ (one decimal) is

Numerical answer type, answer in mL/mol, tolerance 0.2.

Answer: 37.3 mL/mol

b = \frac{R T _{c}}{8 P _{c}} = \frac{0.0821 \times 405}{8 \times 111.5} = \frac{33.251}{892} = 0.03728 L mo l^{- 1} = 37.28 mL mo l^{- 1}

, i.e.

37.3 mL mo l^{- 1}