1.0 g of non - electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.4 K. Find the molar mass of the solute. [Given : Freezing point depression constant of benzene = 5.12 K. kg mol].

(a) 10 g of non-electrolyte solute dissolved in 50 g of benzene lowered the freexing point of benzene by 0.4 K. Find the molar mass of the solute.[Given : Freezing point depression constant of benzene = 5.12 K. kg mor^(-1)] (b) How solubility of a gas in liquid varies with (i) Temperature and (ii) pressure?

If 10.0 g of a non-electrolyte dissolved in 100 g of water lowers the freezing point of water by 1.86^(@)C , the molar mass of the non-electrolyte is ( K_(f)=1.86 Km^(-1) )

0.90g of a non-electrolyte was dissolved in 90 g of benzene. This raised the boiling point of benzene by 0.25^(@)C . If the molecular mass of the non-electrolyte is 100.0 g mol^(-1) , calculate the molar elevation constant for benzene.

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) ]

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) )

By dissolving 13.6 g of a substance in 20 g of water, the freezing point decreased by 3.7^(@)C . Calculate the molecular mass of the substance. (Molal depression constant for water = 1.863K kg mol^(-1))

Two elements A and B form compounds having molecular formula AB_(2) and AB_(4) . When dissolved in 20 g of benzene, 1 g of AB_(2) lowers the freezing point by 2.3 K , whereas 1.0 g of AB_(4) lowers it by 1.3 K . The molar depression constant for benzene is 5.1 K kg mol^(-1) . Calculate the atomic mass of A and B .