The vapour pressure of benzene at`80^(@)C` is lowered by `10mm` by dissoving `2g`of a non-volatile substance in`78g`of benzene .The vapour pressure of pure benzene at `80^(@)C`is `750mm`.The molecular weight of the substance will be:

The vapour pressure of benzene at 80^@C is lowered by 10 mm by dissolving 2g of a non- volatile substance in 78 g of benzene . The vapour pressure of pure benzene at 80^(@)C is 750 mm . The molecular mass of the substance will be _______ g mol ^(-1) ?

Vapour pressure of benzene at 30^(@)C is 121.8 mm. When 15 g of a non-volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute is (mol. Weight of solvent = 78)

Vapour pressure of benzene at 30^(@)C is 121.8 mm. When 15 g of a non-volatile solute is dissolved in 250 g of benzene its vapour pressure decreased to 120.2 mm. The molecular weight of the solute is (mol. Weight of solvent = 78)

64 g non - volatile solute is added to 702 g benzene. The vapour pressure of benzene has decreased from 200 mm of Hg to 180 mm of Hg. Molecular weight of the solute is

64 g non - volatile solute is added to 702 g benzene. The vapour pressure of benzene has decreased from 200 mm of Hg to 180 mm of Hg. Molecular weight of the solute is