1.5 g of a non-volatile non-electrolyte ...

1.5 g of a non-volatile non-electrolyte is dissolved in 50 g of benzene `(K_b=2.5K kg mol^-1)`. The elevation of boiling point of the solution is 0.75K. The molecular weigth of the solute is `gmol^-1` is

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Boiling point of benzene is 353.23 K . When 1 . 80 g of non-volatile solute was dissolved in 90 g of benzene the boiling point is raised to 354. 11 K? Calculate molar mass of solute . ( K_(b) for benzene is 2 . 53 K kg mol^(-1) )

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For a dilute solution containing 2.5 g of a non volatile non electrolyte solution in 100 g of water, the elevation in boiling point at 1 atm pressure is 2^@C . Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (take k b = 0.76 K Kg mol^-1 )

10g of non-volatile solute when dissolved in 100g of benzene raises its boiling point by 1^@ C. What is the molecular mass of the solute. ( k_b for benzene=2.53 K kg mol^-1 )

The boiling point of benzene is 353.23 K. When 1.80g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute.

1.00g of a non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40K. The freezing point depression constant of benzene is 5.12K mol^-1 . Find the molar mass of the solute.

The boiling point of Benzene (C_(6) H_(6)) is 353.23 K. When 1 . 80 g of a non - volatile solute was dissolved in 90 g of C_(6) H_(6) the boiling point is raised to 354 . 11 K. Calculate the molar mass of solute . (Given K_(b) for benzene is 2 . 53 K kg mol^(-1) )

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