Home
Class 12
CHEMISTRY
Calculate the freezing point of an aqueo...

Calculate the freezing point of an aqueous soltuion of non-electrolyte having an osmotic pressure `2.0 atm` at `300 K`. (`K'_(f) = 1.86 K mol^(-1) kg` and `S = 0.0821` litre atm `K^(-1) mol^(-1)`)

A

`-0.15^(@)C`

B

`+0.15^(@)C`

C

`-0.51^(@)C`

D

`-0.17^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
A
`pi = CRT`
`:. C = (pi)/(RT)`
`=(2.0 atm)/((0.0821 L atm K^(-1) mol^(-1))xx(300K))`
`=0.0812 mol L^(-1) =0.0812 M ~~0.0812 m`
`( :'` solution is very dilute)
`Delta T_(f) = K_(f)m = (1.86 Km^(-1)) xx (0.0812m)`
`=0.15K = 0.15^(@)C`
`:.` Freezing point `= (0.00 - 0.15)^(@)C =- 0.15^(@)C`.
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    DINESH PUBLICATION|Exercise REASON ASSERTION TYPE|21 Videos

  • SOLID STATE

    DINESH PUBLICATION|Exercise Brain storming|10 Videos

  • SOME BASIC CONCEPTS OF CHEMISTRY

    DINESH PUBLICATION|Exercise ULTIMATE PREPARATORY PACKAGE|20 Videos

Similar Questions

Explore conceptually related problems

The freezing point of an aqueous solution of non-electrolyte having an osmotic pressure of 2.0 atm at 300 K will be, (K_(f) "for " H_(2)O = 1.86K mol^(-1)g) and R = 0.0821 litre atm K^(-1) mol^(-1) . Assume molarity and molality to be same :

Calculate the freezing point of an aqueous solution of a non-electrolyte having an osmotic pressure of 2.0 atm at 300 K. K_f=1.86 k//m, R=0.0821 L "atm" //k//mol .

The freezing point of a 0.05 molal solution of a non-electrolyte in water is [K_(f)=1.86K//m]

Find the freezing point of a glucose solution whose osmotic pressure at 25^(@)C is found to be 30 atm. K_(f) (water) = 1.86 kg mol^(-1)K .

The freezing poing of an aqueous solution of a non-electrolyte is -0.14^(@)C . The molarity of this solution is [K_(f) (H_(2)O) = 1.86 kg mol^(-1)] :

The freezing point of a 0.05 molal solution of a non-electrolyte in water is: ( K_(f) = 1.86 "molality"^(-1) )

Calculate the freezing point of a solution containing 60 g glucose (Molar mass = 180 g mol^(-1) ) in 250 g of water . ( K_(f) of water = 1.86 K kg mol^(-1) )