Home
Class 12
CHEMISTRY
(a) 31 g of an unknown molecular materia...

(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.

Text Solution

Verified by Experts

(a)`M_(B) = K_(f)W_(B)1000/DeltaT_(f)W_(A)`
`Delta(T)_(f) =``T^(0)_(f) - T_(f)=273-271.14K=1.86K`
`M_(B)= 1.86xx31xx1000/(1.86xx500)=62gmol^(-1)`
Promotional Banner

Topper's Solved these Questions

  • ANNUAL EXAMINATION QUESTION PAPER MARCH 2019

    SUBHASH PUBLICATION|Exercise PART C|8 Videos
  • ANNUAL EXAMINATION QUESTION PAPER MARCH 2018

    SUBHASH PUBLICATION|Exercise PART-D|24 Videos
  • BIOMOLECULES

    SUBHASH PUBLICATION|Exercise QUESTION|77 Videos

Similar Questions

Explore conceptually related problems

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 .

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 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))

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 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

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))

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) )