Home
Class 12
CHEMISTRY
An aqueous solution contaning 5% by weig...

An aqueous solution contaning `5%` by weight of urea and `10%` weight of glucose. What will be its freezing point? `(K'_(f) for H_(2)O "is" 1.86^(circ) mol^(-1) kg)`

Text Solution

Verified by Experts

`Delta T=DeltaT_("urea")+DeltaT_("glucose")` ,
`DeltaT=(1000xx1.86xx5)/(60xx85)+(1000xx1.86xx10)/(180xx85)`
`=1.824+1.216`
=3.04
Freezing point `=0-3.04=-3.04^(@)C`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    AAKASH INSTITUTE|Exercise TRY YOURSELF|14 Videos

  • SOLUTIONS

    AAKASH INSTITUTE|Exercise ASSIGMENT (SECTION-A)|50 Videos

  • REDOX REACTIONS

    AAKASH INSTITUTE|Exercise ASSIGNMENT SECTION - D|10 Videos

  • SOME BASIC CONCEPT OF CHEMISTRY

    AAKASH INSTITUTE|Exercise ASSIGNMENT( SECTION - D) Assertion-Reason Type Questions|15 Videos

Similar Questions

Explore conceptually related problems

An aqueous solution contains 5% by mass of urea and 10% by mass of glucose . If K_f for water is 1.86 " k mol " ^(-1) , the freezing point of the solution is

An aqueous solution contain 3% and 1.8% by mass. Urea and glucose respectively. What is the freezing point of solution ? ( K_(f)=1.86^(@)C//m )

The freezing point of aqueous solution that contains 5% by mass urea. 1.0% by mass KCl and 10% by mass of glucose is: (K_(f) H_(2)O = 1.86 K "molality"^(-1))