Question

An aqueous solution freezes at -0.186c kf =1.86 kb=0.512.what is the elevation on boiling point

Answer 1

Answer:

hope this will help you

Answer 2

Given:

An aqueous solution freezes at -0.186c kf =1.86 kb=0.512.

To Find:

The elevation on boiling point.

Solution:

To find the elevation in boiling point we will follow the following steps:

As we know,

Elevation in boiling point = kbm

T2-T1 = kbm

Here, m is molality, expressed in mol per kg.

Also,

Depression in freezing point = kfm = T3-T4

kf and kb are constant and depend on the solvent.

T is denoting temperature.

From the above information, we get,

Depression in freezing point = 0.186°c

Now,

0.186 = 1.86×molality

So,

$Molality = \frac{0.186 }{1.86} = \: 0.1$

Elevation in boiling point = T2- T1

T2- T1 = kbm

= 0.152×0.1

= 0.0152°c

Henceforth, the elevation of the boiling point is 0.0152°c.