JEE MAIN - Chemistry (2017 - 9th April Morning Slot - No. 18)
A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2
and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of Cl = 35.5 g mol−1)
0.162
0.675
0.325
0.486
Explanation
Mole fractions can be calculated as
$${n_{C{H_2}C{l_2}}} = {{8.5} \over {8.5}} = 0.1$$ and $${n_{CHC{l_3}}} = {{11.9} \over {119.5}} = 0.1$$
We know that
$${p_{total}} = p_{C{H_2}C{l_2}}^o \times \,{x_{C{H_2}C{l_2}}} + p_{CHC{l_3}}^o \times \,{x_{CHC{l_3}}}$$
$$ = 415 \times 0.10 + 200 \times 0.1 = 61.5$$
Mole fraction of CHCl3 in vapour form can be calculated as
$${p_{total}} = p_{CHC{l_3}}^o \times \,{x_{CHC{l_3}}}$$
$$61.5 = 200 \times \,{x_{CHC{l_3}}}$$
$${x_{CHC{l_3}}} = {{61.5} \over {200}} = 0.3075$$
Comments (0)
