Relative Lowering of Vapour Pressure
CHXII02:SOLUTIONS

319347 In an experiment, 1g of a non-volatile solute was dissolved in 100g of acetone (mol. mass = 58) at 298K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of acetone =195 mm Hg)

1 25.24
2 35.24
3 44.66
4 55.24
CHXII02:SOLUTIONS

319348 The solubility of a specific non-volatile salt is 4 g in 100 g of water at \(25^{\circ} \mathrm{C}\). If \(2.0 \mathrm{~g}, 4.0 \mathrm{~g}\) and 6.0 g of the salt is added to 100 g of water at \(25^{\circ} \mathrm{C}\), in system \(\mathrm{X}, \mathrm{Y}\) and Z , respectively. The vapour pressure would be in the order

1 \(\mathrm{X} < \mathrm{Y} < \mathrm{Z}\)
2 \({\rm{X}} > {\rm{Y}} > {\rm{Z}}\)
3 \({\rm{Z}} > {\rm{X}} = {\rm{Y}}\)
4 \({\rm{X}} > {\rm{Y}} = {\rm{Z}}\)
CHXII02:SOLUTIONS

319349 What is vapour pressure of a solution containing \(1 \mathrm{~mol}\) of a non - volatile solute in \(36 \mathrm{~g}\) of water? \(\left(\mathrm{P}_{1}^{0}=400 \mathrm{~mm} \mathrm{Hg}\right)\)

1 \(334 \mathrm{~mm} \mathrm{Hg}\)
2 \(267 \mathrm{~mm} \mathrm{Hg}\)
3 \(240 \mathrm{~mm} \mathrm{Hg}\)
4 \(284 \mathrm{~mm} \mathrm{Hg}\)
CHXII02:SOLUTIONS

319350 At a certain temperature vapour pressure of pure water is \(3000 \mathrm{Nm}^{-2} .5 \mathrm{~g}\) of non-electrolyte and non-volatile solute is added to 100 g of water. Vapour pressure of the solution is \(2985\,\,{\rm{N}}{{\rm{m}}^{ - 2}}\). Assume that it is a dilute solution, find the molar mass of the solute.

1 90
2 180
3 200
4 270
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
CHXII02:SOLUTIONS

319347 In an experiment, 1g of a non-volatile solute was dissolved in 100g of acetone (mol. mass = 58) at 298K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of acetone =195 mm Hg)

1 25.24
2 35.24
3 44.66
4 55.24
CHXII02:SOLUTIONS

319348 The solubility of a specific non-volatile salt is 4 g in 100 g of water at \(25^{\circ} \mathrm{C}\). If \(2.0 \mathrm{~g}, 4.0 \mathrm{~g}\) and 6.0 g of the salt is added to 100 g of water at \(25^{\circ} \mathrm{C}\), in system \(\mathrm{X}, \mathrm{Y}\) and Z , respectively. The vapour pressure would be in the order

1 \(\mathrm{X} < \mathrm{Y} < \mathrm{Z}\)
2 \({\rm{X}} > {\rm{Y}} > {\rm{Z}}\)
3 \({\rm{Z}} > {\rm{X}} = {\rm{Y}}\)
4 \({\rm{X}} > {\rm{Y}} = {\rm{Z}}\)
CHXII02:SOLUTIONS

319349 What is vapour pressure of a solution containing \(1 \mathrm{~mol}\) of a non - volatile solute in \(36 \mathrm{~g}\) of water? \(\left(\mathrm{P}_{1}^{0}=400 \mathrm{~mm} \mathrm{Hg}\right)\)

1 \(334 \mathrm{~mm} \mathrm{Hg}\)
2 \(267 \mathrm{~mm} \mathrm{Hg}\)
3 \(240 \mathrm{~mm} \mathrm{Hg}\)
4 \(284 \mathrm{~mm} \mathrm{Hg}\)
CHXII02:SOLUTIONS

319350 At a certain temperature vapour pressure of pure water is \(3000 \mathrm{Nm}^{-2} .5 \mathrm{~g}\) of non-electrolyte and non-volatile solute is added to 100 g of water. Vapour pressure of the solution is \(2985\,\,{\rm{N}}{{\rm{m}}^{ - 2}}\). Assume that it is a dilute solution, find the molar mass of the solute.

1 90
2 180
3 200
4 270
CHXII02:SOLUTIONS

319347 In an experiment, 1g of a non-volatile solute was dissolved in 100g of acetone (mol. mass = 58) at 298K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of acetone =195 mm Hg)

1 25.24
2 35.24
3 44.66
4 55.24
CHXII02:SOLUTIONS

319348 The solubility of a specific non-volatile salt is 4 g in 100 g of water at \(25^{\circ} \mathrm{C}\). If \(2.0 \mathrm{~g}, 4.0 \mathrm{~g}\) and 6.0 g of the salt is added to 100 g of water at \(25^{\circ} \mathrm{C}\), in system \(\mathrm{X}, \mathrm{Y}\) and Z , respectively. The vapour pressure would be in the order

1 \(\mathrm{X} < \mathrm{Y} < \mathrm{Z}\)
2 \({\rm{X}} > {\rm{Y}} > {\rm{Z}}\)
3 \({\rm{Z}} > {\rm{X}} = {\rm{Y}}\)
4 \({\rm{X}} > {\rm{Y}} = {\rm{Z}}\)
CHXII02:SOLUTIONS

319349 What is vapour pressure of a solution containing \(1 \mathrm{~mol}\) of a non - volatile solute in \(36 \mathrm{~g}\) of water? \(\left(\mathrm{P}_{1}^{0}=400 \mathrm{~mm} \mathrm{Hg}\right)\)

1 \(334 \mathrm{~mm} \mathrm{Hg}\)
2 \(267 \mathrm{~mm} \mathrm{Hg}\)
3 \(240 \mathrm{~mm} \mathrm{Hg}\)
4 \(284 \mathrm{~mm} \mathrm{Hg}\)
CHXII02:SOLUTIONS

319350 At a certain temperature vapour pressure of pure water is \(3000 \mathrm{Nm}^{-2} .5 \mathrm{~g}\) of non-electrolyte and non-volatile solute is added to 100 g of water. Vapour pressure of the solution is \(2985\,\,{\rm{N}}{{\rm{m}}^{ - 2}}\). Assume that it is a dilute solution, find the molar mass of the solute.

1 90
2 180
3 200
4 270
CHXII02:SOLUTIONS

319347 In an experiment, 1g of a non-volatile solute was dissolved in 100g of acetone (mol. mass = 58) at 298K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of acetone =195 mm Hg)

1 25.24
2 35.24
3 44.66
4 55.24
CHXII02:SOLUTIONS

319348 The solubility of a specific non-volatile salt is 4 g in 100 g of water at \(25^{\circ} \mathrm{C}\). If \(2.0 \mathrm{~g}, 4.0 \mathrm{~g}\) and 6.0 g of the salt is added to 100 g of water at \(25^{\circ} \mathrm{C}\), in system \(\mathrm{X}, \mathrm{Y}\) and Z , respectively. The vapour pressure would be in the order

1 \(\mathrm{X} < \mathrm{Y} < \mathrm{Z}\)
2 \({\rm{X}} > {\rm{Y}} > {\rm{Z}}\)
3 \({\rm{Z}} > {\rm{X}} = {\rm{Y}}\)
4 \({\rm{X}} > {\rm{Y}} = {\rm{Z}}\)
CHXII02:SOLUTIONS

319349 What is vapour pressure of a solution containing \(1 \mathrm{~mol}\) of a non - volatile solute in \(36 \mathrm{~g}\) of water? \(\left(\mathrm{P}_{1}^{0}=400 \mathrm{~mm} \mathrm{Hg}\right)\)

1 \(334 \mathrm{~mm} \mathrm{Hg}\)
2 \(267 \mathrm{~mm} \mathrm{Hg}\)
3 \(240 \mathrm{~mm} \mathrm{Hg}\)
4 \(284 \mathrm{~mm} \mathrm{Hg}\)
CHXII02:SOLUTIONS

319350 At a certain temperature vapour pressure of pure water is \(3000 \mathrm{Nm}^{-2} .5 \mathrm{~g}\) of non-electrolyte and non-volatile solute is added to 100 g of water. Vapour pressure of the solution is \(2985\,\,{\rm{N}}{{\rm{m}}^{ - 2}}\). Assume that it is a dilute solution, find the molar mass of the solute.

1 90
2 180
3 200
4 270
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here