Home
Class 12
CHEMISTRY
Freezing point of an aqueous solution is...

Freezing point of an aqueous solution is `-0.186^(@)C`. If the values of `K_(b)andK_(f)` of water are respectively 0.52 K kg `mol^(-1)` and 1.86 K kg `mol^(-1)` then the elevation of boiling point of the solution in K is

A

`0.52`

B

`1.04`

C

`1.34`

D

`0.052`

Text Solution

Verified by Experts

The correct Answer is:
D
`because` Freezing point of the aqueous solution
`=-0.186^(@)C`
`therefore` Depression in freezing point,
`DeltaT_(f)=0^(@)C-(-0.186^(@)C)`
`=+0.186^(@)C=+0.186K`
We know that, `DeltaT_(f)=K_(f).m …(i)`
and elevation in boiling piont,
`DeltaT_(b)=K_(b).m ...(ii)`
Eq. (i) Eq. (ii) gives
`(DeltaT_(f))/(DeltaT_(b))=(K_(f))/(K_(b))`
`implies (0.186K)/(DeltaT_(b))=(1.86K kg mol^(-1))/(0.52K kg mol^(-1))`
`DeltaT_(b)=(0.186xx0.52)/(1.86)`
`=0.052K`
Promotional Banner

Similar Questions

Explore conceptually related problems

If boiling point of an aqueous solution is 100.1^(@)C , its freezing point is ( I_(f) of water =80 cal/g I_(v) of water =540 cal/g)

0.2 molal aqueous solution of acid HX is 20% ionised. K_f="1.86 K molal"^(-1) . The freezing point of the solution is very close to:

The molality of an aqueous solution of methanol (CH_3OH) is 1.44 mol kg ""^(-1) . The mole fraction of methanol in the solution will be

An aqueous solution solution of a non-volatile solute boils at 373.053K. Find its freezing point. For water, K_b=0.52K kg mol^(-1), K_f = 1.86K kg mol^(-1) Boiling point =373K, freezing point of water =273K

Elevation in boiling point of an aqueous glucose solution is 0.6, K_b for water is 0.52 K molal ""^(-1). The mole fraction of glucose in the solution is:

An aqueous solution of glucose boils at 100.01^@C. The molal elevation constant for water is "0.5 K mol"^(-1)" kg". What is the number of glucose molecules in the solution containing 100 g water?

An aqueous solution freezes at -0.186^@C (K_f=1.86 ,K_b, =0.512^@). What is the elevation in boiling point?