JEE MAIN - Chemistry (2015 (Offline) - No. 1)
The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol-1) of the substance is:
64
128
488
32
Explanation
Using relation,
$${{{P^ \circ } - {P_s}} \over {{P_s}}} = {{{w_2}{M_1}} \over {{w_1}{M_2}}}$$
where $${w_1},$$ $${M_1} = $$ mass in $$g$$
and mol. mass of solvent
$${w_2},{M_2} = $$ mass in $$g$$
and mol. mass of solute
Let $${M_2} = x$$
$${P^ \circ } = 185\,\,torr;\,\,{P_s} = 183\,torr$$
$${{185 - 183} \over {183}} = {{1.2 \times 58} \over {100x}}$$
(Mol. mass of acetone $$=58$$)
$$x=64$$
$$\therefore$$ $$\,\,\,\,\,$$ Molar mass of substance $$=64$$
$${{{P^ \circ } - {P_s}} \over {{P_s}}} = {{{w_2}{M_1}} \over {{w_1}{M_2}}}$$
where $${w_1},$$ $${M_1} = $$ mass in $$g$$
and mol. mass of solvent
$${w_2},{M_2} = $$ mass in $$g$$
and mol. mass of solute
Let $${M_2} = x$$
$${P^ \circ } = 185\,\,torr;\,\,{P_s} = 183\,torr$$
$${{185 - 183} \over {183}} = {{1.2 \times 58} \over {100x}}$$
(Mol. mass of acetone $$=58$$)
$$x=64$$
$$\therefore$$ $$\,\,\,\,\,$$ Molar mass of substance $$=64$$
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