15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at - `0.34^(@)`C. what is molar mass of the substance ? `(K_(f)" for water " = 1.86 k kg " mol"^(-1))`

(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.

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 .

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))

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) ).

Calculate the molar mass of a substance 1 g of which when dissolved in 100 g of water gave a solution boiling at 100.1^(@)C at a pressure of 1 atm (K_(b) for water = 0.52 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))

What do you understand by the term that K_(f) for water is 1.86 K kg mol^(-1) ?

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) )