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

a) 24 g of a non-volatile, non-electrolyte solute is added to 600 g of water. The boiling point of the resulting solution is 373.35K. Calculate the molar mass of the solute ("Given boiling point of pure water = 373 K and Kb for water=0.52 K kg mol"^(-1)) .

6 grams of a substance 'A' dissolved in 100 g of water freezes at - 0.93^(@)C . The molecular mass of 'A' is ( K_(f)=1.86 Km^(-1) ).

A solution containing 12.5 g of non-electrolyte solution in 175g of water gave a boiling point elevation of 0.7 k. calculate the molar mass of the solute if K_(b) for water is 0.52 K kg mol^(-1) .

A solution containg 12 g of a non-electrolyte substance in 52 g of water gave boiling point elevation of 0.40 K . Calculate the molar mass of the substance. (K_(b) for water = 0.52 K kg mol^(-1))

Answer any three of the following questions. a) 31 g of an unknown molecular material is dissolved in 500 g of water. The resulting solution freezes at 271.14 K. Calculate the molar mass of the material. Given : K_t for water =1.86"KK gmol1" T_(t)^(@) of water =273K .

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) 31 g of an unknown molecular material is dissolved in 500 g of water. The resulting solution freezes at 27.14 K. Calculate the molar mass of the material. [Given: K, for water 1.86 Kgmol^(-1) , T_(f)^(0) of water=273К]. (b) What is reverse osmosis? Mention its use.

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) ]