Question

Kb of water = 0.52 kkg mol–1. The B.P of a solution of 0.1 mol of cane sugar in 200g of water is:
1) 99.74oC
2) 100.52oC
3) 99.48oC
4) 100.26oC
pls answer with explanation .. will mrk as brainliest

Answer 1

Answer:

1)

Explanation:

is the correct answer and please mark as brainlest answer die in the future of the interest rate r reader TM

Answer 2

Answer:

4) 100.26°C

Explanation:

Cane Sugar is solute, water is solvent

ΔTb = $\frac{Kb 1000 w2 }{m2w1}$

Where,

ΔTb is the Elevation in Boiling Point

Kb is the boiling point elevation constant or molal elevation constant (ebullioscopic constant)

w2 represents the weight of the solute

m2 represents the molar mass of the solute

w1 represents the weight of the solvent

We know,

Weight/Mass = No. of Moles

So,

n2 = w2/m2

In the question cane sugar (the solute) has 0.1 mol

So instead of w2/m2, we can subtitute the number of moles in the equation.

ΔTb = $\frac{Kb 1000 n2 }{w1}$

Now substituting,

= 0.52 X 1000 X 0.1 / 200

=  0.26

This is the elevation in boiling point.

The normal boiling point as we all know, is 100°C

Elevation represents increase,

hence

= 100°C + 0.26°C

= 100.26°C