Question

# A solution is prepared by mixing 8.5 g of $$CH_2Cl_2$$ and 11.95 g of $$CHCl_3$$. If vapour pressure of $$CH_2Cl_2$$ and $$CHCl_3$$ at 298 K are 415 and 200 mmHg respectively, the mole fraction of $$CHCl_3$$ in vapour form is:[Molar mass of C1=35.5 g $$mol^{-1}$$]

A
0.325
B
0.486
C
0.675
D
0.162

Solution

## The correct option is A 0.325Given$$P_0=$$ Vapor pressure = 200 mm of HgMass of solute (CH$$_2$$Cl$$_2$$)=8.5 gMass of solvent (CHCl$$_3$$)=11.95 gMole fraction in solution phase can be calculated as:$$X_{CH_2Cl_2}=\dfrac{(8.5/85)}{(8.5/85)+(11.95/119.5)}=0.5$$$$X_{CHCl_3}=\dfrac{(11.95/119.5)}{(8.5/85)+(11.95/119.5)}=0.5$$Mole fraction in vapor phase is given by the formula:$$Y_{CHCl_3}=\dfrac{P^o.X_{CHCl_3}}{P^o.X_{CHCl_3}+P^o.X_{CH_2Cl_2}}=\dfrac{200\times0.5}{200\times0.5+415\times0.5}=0.325$$Hence, the correct option is $$A$$Chemistry

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