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

The boiling point of benzene is 353.23 K when 1.80 g of a non-volatile, non-ionising solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute. " " [Given K_(b) for benzene = 2.53 K kg mol^(-1) ]

(a) The boiling point of benzene is 353.23K. When 1.80g of a non-volatile,non-ionisable solute was dissolved in 90g benzene, the boiling point raised to 354.11K. Calculate molarass of the solute.[Kb for benzene =2.53K kg mol-1]. (b)Define :(i) molality of a solution. (ii) Isotonic solutions.

The boiling point of benzene is 353.23 K . When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to 354.11 K . Calculate the molar mass of the solute. ( K_(b) for benzene is 2.53 K kg mol^(-1) )

Vapour pressure of benzene is 200 mm of Hg. When 2 gram of a non-volatile solute dissolved in 78 gram benzene, benzene has vapour pressure of 195 mm of Hg. Calculate the molar mass of the solute. [Molar mass of benzene is 78 g/ mol^(-1) ]

Addition of non-volatile solute always increases the boiling point of the solution. Why?

A solution containing 18g of non - volatile non - electrolyte solute is dissolved in 200g of water freezes at 272.07K. Calculate the molecular mass of solute. Given K_(f)=1.86kg//mol and freezing point of water = 273K

The boiling point of a solvent containing a non-volatile solute

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