JEE Advance - Chemistry (1992 - No. 12)
At 27oC, hydrogen is leaked through a tiny hole into a vessel for 20 minutes. Another unknown gas at the same temperature and pressure as that of H2 is leaked through the same hole for 20 minutes After the effusion of the gases the mixture exerts a pressure of 6 atmosphere. The hydrogen content of the mixture is 0.7 mole. If the volume of ther container is 3 litres, what is the molecular weight of the unknown gas?
Risposta
1033
Spiegazione
$${p_{{H_2}}} + {p_g} = 6.0$$ atm
where pg is the pressure exerted by the unknown gas.
$${p_{{H_2}}} = {{nRT} \over V} = {{0.7 \times 0.0821 \times 300} \over 3} = 5.747$$ atm.
$$\therefore$$ $${p_g} = 6.0 - 5.747 = 0.253$$ atm.
Number of moles of unknown gas
$$ = {{{p_g}\,.\,V} \over {RT}} = {{0.253 \times 3} \over {0.0821 \times 300}} = 0.0308$$
Rate of effusion of
$${H_2} = {{0.7} \over {20}} = 0.035$$ mol min$$-$$1
Rate of effusion of unknown gas
$$ = {{0.0308} \over {20}} = 0.00154$$ mol min$$-$$1
According to Graham's Law of effusion
$${{{M_g}} \over {{M_{{H_2}}}}} = {{{{({r_{{H_2}}})}^2}} \over {{{({r_g})}^2}}} = {{{{(0.035)}^2}} \over {{{(0.00154)}^2}}} = 516.5$$
$$\therefore$$ $${M_g} = 516.5 \times 2 = 1033$$ g mol$$-$$1