Vapour pressure of chloroform (CHCl_3) and dichloromethane (CH_2 Cl_2) at 25^ C are 200 ~mm Hg and 415 ~mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 ~g of CHCl_3 and 4 O g of CH_2 Cl_2 at

Correct answer: C. 347.9 mmHg Molar mass of CH_2 Cl_2=12 × 1+1 × 2+35.5 × 2=85 ~g ~mol^ -1 Molar mass of CHCl_3=12 × 1+1 × 1+35.5 × 3=119.5 ~g ~mol^ -1 Moles of CH_2 Cl_2=40 ~g / 85 ~g ~mol^ -1 =0.47 ~mol Moles of CHCl_3=25.5 ~g / 119.5 ~g ~mol^ -1 =0.213 ~mol Total number of moles =0.47+0.213=0.683 ~mol Mole fraction of component 2 =0.47 ~mol / 0.683 ~mol= 0.688 Mole fraction of component 1 =1.00-0.688=0.312 We know that: aligned & P_ T =p_ 1 ^0+(p_ 2 ^0-p_ 1 ^0) x_ 2 & =200+(415-200) ×0.688 & =200+147.9 & =347.9 ~mm Hg aligned

NEET 2012

Exam: NEET
Year: 2012
Subject: Chemistry
Chapter: Solutions

NEET 2012 · Chemistry · Solutions

Vapour pressure of chloroform (CHCl_3) and dichloromethane (CH_2...

Vapour pressure of chloroform

\left(\mathrm{CHCl}_3\right)

and dichloromethane

\left(\mathrm{CH}_2 \mathrm{Cl}_2\right)

25^{\circ} \mathrm{C}

are

200 \mathrm{~mm} \mathrm{Hg}

and

415 \mathrm{~mm} \mathrm{Hg}

respectively. Vapour pressure of the solution obtained by mixing

25.5 \mathrm{~g}

\mathrm{CHCl}_3 and 4 \mathrm{O} \mathrm{g} of \mathrm{CH}_2 \mathrm{Cl}_2

at the same temperature will be: [Molecular mass of

\mathrm{CHCl}_3=119.5 \mathrm{~g} / \mathrm{mol}

and molecular mass of

\mathrm{CH}_2 \mathrm{Cl}_2=85 \mathrm{~g} / \mathrm{mol}

]

Options

A $173.9 \mathrm{mmHg}$
B $615.0 \mathrm{mmHg}$
C $347.9 \mathrm{mmHg}$
D $28.5 \mathrm{mmHg}$

Correct answer

C. $347.9 \mathrm{mmHg}$

Step-by-step solution

Molar mass of

\mathrm{CH}_2 \mathrm{Cl}_2=12 \times 1+1 \times 2+35.5 \times 2=85 \mathrm{~g} \mathrm{~mol}^{-1}

Molar mass of

\mathrm{CHCl}_3=12 \times 1+1 \times 1+35.5 \times 3=119.5 \mathrm{~g} \mathrm{~mol}^{-1}

Moles of

\mathrm{CH}_2 \mathrm{Cl}_2=40 \mathrm{~g} / 85 \mathrm{~g} \mathrm{~mol}^{-1}=0.47 \mathrm{~mol}

Moles of

\mathrm{CHCl}_3=25.5 \mathrm{~g} / 119.5 \mathrm{~g} \mathrm{~mol}^{-1}=0.213 \mathrm{~mol}

Total number of moles

=0.47+0.213=0.683 \mathrm{~mol}

Mole fraction of component 2 =0.47

\mathrm{~mol}

/ 0.683

\mathrm{~mol}=

0.688 Mole fraction of component 1 =1.00-0.688=0.312 We know that:

\begin{aligned}& \mathrm{P}_{\mathrm{T}}=\mathrm{p}_{1}^0+\left(\mathrm{p}_{2}^0-\mathrm{p}_{1}^0\right) \mathrm{x}_{2} \\ & =200+(415-200) \times0.688 \\ & =200+147.9 \\ & =347.9 \mathrm{~mm} \mathrm{Hg} \end{aligned}

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