(B) The ideal gas law is valid only if the volume occupied by the gas molecules is negligible compared to the total volume of the container.
$1$. Calculate the number of moles of $H_2$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.5 \, g}{2 \, g/mol} = 0.25 \, mol$.
$2$. Calculate the total number of $H_2$ molecules $(N)$: $N = n \times N_A = 0.25 \times 6.023 \times 10^{23} \approx 1.506 \times 10^{23}$ molecules.
$3$. Calculate the volume of a single $H_2$ molecule $(v_m)$: $v_m = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (10^{-10} \, m)^3 \approx 4.19 \times 10^{-30} \, m^3$.
$4$. Calculate the total volume occupied by the molecules $(V_{mol})$: $V_{mol} = N \times v_m = 1.506 \times 10^{23} \times 4.19 \times 10^{-30} \approx 6.31 \times 10^{-7} \, m^3$.
$5$. Calculate the final volume of the chamber $(V_f)$ using Boyle's Law $(P_i V_i = P_f V_f)$:
$V_i = (3 \, cm)^3 = 27 \, cm^3 = 27 \times 10^{-6} \, m^3$.
$P_i = 1 \, atm$,$P_f = 100 \, atm$.
$V_f = \frac{P_i V_i}{P_f} = \frac{1 \times 27 \times 10^{-6}}{100} = 2.7 \times 10^{-7} \, m^3$.
$6$. Comparison: The volume occupied by the molecules $(6.31 \times 10^{-7} \, m^3)$ is greater than the total volume of the container $(2.7 \times 10^{-7} \, m^3)$. Since the molecular volume is not negligible,the ideal gas law is $NOT$ justified.