Addition of 0.643 'g of a compound to '50 mL' of 'benzene (density 0.879 g/mL) lowers the freez ing point from '5.51^circ C' to '5.03^circ C'. If 'K' for benzene is '5.12 ~K' kg mol -1 Calculate the molar mass of the compound?

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

Two elements 'A' and 'B' form compounds having formula 'AB2' and 'AB_4' .When dissolved in '20 g' of benzene '(C_6 H_6), 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 atomic masses of A and B.

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

A compound has the empirical formula C_(10)H_(8) Fe . A solution of 0.26 gof the compound in 11.2 g of benzene (C_6H_6) boils at 80.26^@C. The boiling point of benzene is 80.10^@C, K_b" is 2.53"^@C"/molal". What is the molecular formula of the compound?

The density of gas was found to be 2.92 g L^(-1) at 27° C and 2.0 atm . Calculate the molar mass of the gas.

34.05 mL' of phosphorus vapour weighs '0.0625 g' at '546^degree C' and '0.1' bar pressure. What is the molar mass of phosphorus?

A '1.28 g' sample of a colourless liquid was vapourised in a '250 ml' flask at '121^circ C' and '786 mm Hg'. What is the molar mass of this substance?

0.48g of a substance was dissolved in 10.6 g C_6H_6. The freezing point of benzene was lowered by 1.8^@C. Calculate molar mass of the substance. Molecular depression constant for benzene is "5K kg mol"^(-1) .

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