Relative Lowering of Vapour Pressure
CHXII02:SOLUTIONS

319343 What is vapour pressure of solution containing \(1.8 \mathrm{~g}\) glucose in \(16.2 \mathrm{~g}\) water?
\(\left(\mathrm{P}_{1}^{0}=24 \mathrm{mmHg}\right.\) and Molar mass of glucose \(=180 \mathrm{~g} \mathrm{~mol}^{-1}\) )

1 \(18.1 \mathrm{mmHg}\)
2 \(15.7 \mathrm{~mm} \mathrm{Hg}\)
3 \(12.4 \mathrm{~mm} \mathrm{Hg}\)
4 \(23.8 \mathrm{mmHg}\)
CHXII02:SOLUTIONS

319344 The vapour pressure of pure \(\mathrm{CHCl}_{3}\) and \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are 200 and 41.5 atm respectively. The weight of \(\mathrm{CHCl}_{3}\) and \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are respectively \(11.9 \mathrm{~g}\) and \(17 \mathrm{~g}\). The vapour pressure of solution will be

1 80.5
2 79.5
3 94.3
4 105.5
CHXII02:SOLUTIONS

319345 Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y ?

1 20
2 50
3 100
4 200
CHXII02:SOLUTIONS

319346 Vapour pressure of \({\rm{CC}}{{\rm{l}}_{\rm{4}}}{\mkern 1mu} {\mkern 1mu} {\rm{at}}{\mkern 1mu} {\mkern 1mu} {\rm{2}}{{\rm{5}}^{\rm{^\circ }}}{\rm{C}}\) is 143 mm of Hg. 0.5 gm of a non-volatile solute (molecular weight = 65) is dissolved in 100 ml \({\rm{CC}}{{\rm{l}}_{\rm{4}}}\). Find the vapour pressure of the solution (Density of \({\rm{CC}}{{\rm{l}}_{\rm{4}}}{\rm{ = 1}}{\rm{.58}}{\mkern 1mu} {\mkern 1mu} {\rm{g/c}}{{\rm{m}}^{\rm{3}}}\))

1 143.99 mm
2 94.39 mm
3 199.34 mm
4 141.93 mm
CHXII02:SOLUTIONS

319343 What is vapour pressure of solution containing \(1.8 \mathrm{~g}\) glucose in \(16.2 \mathrm{~g}\) water?
\(\left(\mathrm{P}_{1}^{0}=24 \mathrm{mmHg}\right.\) and Molar mass of glucose \(=180 \mathrm{~g} \mathrm{~mol}^{-1}\) )

1 \(18.1 \mathrm{mmHg}\)
2 \(15.7 \mathrm{~mm} \mathrm{Hg}\)
3 \(12.4 \mathrm{~mm} \mathrm{Hg}\)
4 \(23.8 \mathrm{mmHg}\)
CHXII02:SOLUTIONS

319344 The vapour pressure of pure \(\mathrm{CHCl}_{3}\) and \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are 200 and 41.5 atm respectively. The weight of \(\mathrm{CHCl}_{3}\) and \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are respectively \(11.9 \mathrm{~g}\) and \(17 \mathrm{~g}\). The vapour pressure of solution will be

1 80.5
2 79.5
3 94.3
4 105.5
CHXII02:SOLUTIONS

319345 Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y ?

1 20
2 50
3 100
4 200
CHXII02:SOLUTIONS

319346 Vapour pressure of \({\rm{CC}}{{\rm{l}}_{\rm{4}}}{\mkern 1mu} {\mkern 1mu} {\rm{at}}{\mkern 1mu} {\mkern 1mu} {\rm{2}}{{\rm{5}}^{\rm{^\circ }}}{\rm{C}}\) is 143 mm of Hg. 0.5 gm of a non-volatile solute (molecular weight = 65) is dissolved in 100 ml \({\rm{CC}}{{\rm{l}}_{\rm{4}}}\). Find the vapour pressure of the solution (Density of \({\rm{CC}}{{\rm{l}}_{\rm{4}}}{\rm{ = 1}}{\rm{.58}}{\mkern 1mu} {\mkern 1mu} {\rm{g/c}}{{\rm{m}}^{\rm{3}}}\))

1 143.99 mm
2 94.39 mm
3 199.34 mm
4 141.93 mm
CHXII02:SOLUTIONS

319343 What is vapour pressure of solution containing \(1.8 \mathrm{~g}\) glucose in \(16.2 \mathrm{~g}\) water?
\(\left(\mathrm{P}_{1}^{0}=24 \mathrm{mmHg}\right.\) and Molar mass of glucose \(=180 \mathrm{~g} \mathrm{~mol}^{-1}\) )

1 \(18.1 \mathrm{mmHg}\)
2 \(15.7 \mathrm{~mm} \mathrm{Hg}\)
3 \(12.4 \mathrm{~mm} \mathrm{Hg}\)
4 \(23.8 \mathrm{mmHg}\)
CHXII02:SOLUTIONS

319344 The vapour pressure of pure \(\mathrm{CHCl}_{3}\) and \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are 200 and 41.5 atm respectively. The weight of \(\mathrm{CHCl}_{3}\) and \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are respectively \(11.9 \mathrm{~g}\) and \(17 \mathrm{~g}\). The vapour pressure of solution will be

1 80.5
2 79.5
3 94.3
4 105.5
CHXII02:SOLUTIONS

319345 Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y ?

1 20
2 50
3 100
4 200
CHXII02:SOLUTIONS

319346 Vapour pressure of \({\rm{CC}}{{\rm{l}}_{\rm{4}}}{\mkern 1mu} {\mkern 1mu} {\rm{at}}{\mkern 1mu} {\mkern 1mu} {\rm{2}}{{\rm{5}}^{\rm{^\circ }}}{\rm{C}}\) is 143 mm of Hg. 0.5 gm of a non-volatile solute (molecular weight = 65) is dissolved in 100 ml \({\rm{CC}}{{\rm{l}}_{\rm{4}}}\). Find the vapour pressure of the solution (Density of \({\rm{CC}}{{\rm{l}}_{\rm{4}}}{\rm{ = 1}}{\rm{.58}}{\mkern 1mu} {\mkern 1mu} {\rm{g/c}}{{\rm{m}}^{\rm{3}}}\))

1 143.99 mm
2 94.39 mm
3 199.34 mm
4 141.93 mm
CHXII02:SOLUTIONS

319343 What is vapour pressure of solution containing \(1.8 \mathrm{~g}\) glucose in \(16.2 \mathrm{~g}\) water?
\(\left(\mathrm{P}_{1}^{0}=24 \mathrm{mmHg}\right.\) and Molar mass of glucose \(=180 \mathrm{~g} \mathrm{~mol}^{-1}\) )

1 \(18.1 \mathrm{mmHg}\)
2 \(15.7 \mathrm{~mm} \mathrm{Hg}\)
3 \(12.4 \mathrm{~mm} \mathrm{Hg}\)
4 \(23.8 \mathrm{mmHg}\)
CHXII02:SOLUTIONS

319344 The vapour pressure of pure \(\mathrm{CHCl}_{3}\) and \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are 200 and 41.5 atm respectively. The weight of \(\mathrm{CHCl}_{3}\) and \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are respectively \(11.9 \mathrm{~g}\) and \(17 \mathrm{~g}\). The vapour pressure of solution will be

1 80.5
2 79.5
3 94.3
4 105.5
CHXII02:SOLUTIONS

319345 Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y ?

1 20
2 50
3 100
4 200
CHXII02:SOLUTIONS

319346 Vapour pressure of \({\rm{CC}}{{\rm{l}}_{\rm{4}}}{\mkern 1mu} {\mkern 1mu} {\rm{at}}{\mkern 1mu} {\mkern 1mu} {\rm{2}}{{\rm{5}}^{\rm{^\circ }}}{\rm{C}}\) is 143 mm of Hg. 0.5 gm of a non-volatile solute (molecular weight = 65) is dissolved in 100 ml \({\rm{CC}}{{\rm{l}}_{\rm{4}}}\). Find the vapour pressure of the solution (Density of \({\rm{CC}}{{\rm{l}}_{\rm{4}}}{\rm{ = 1}}{\rm{.58}}{\mkern 1mu} {\mkern 1mu} {\rm{g/c}}{{\rm{m}}^{\rm{3}}}\))

1 143.99 mm
2 94.39 mm
3 199.34 mm
4 141.93 mm