Question

(1) Molal elevation constant kb for water is 0.52 K/m. 0.2 molal solution of glucose
will boil at
(a)100.52°C
(b) 100.052° C
(c) 101.04°C
(d) 100.104°C
answer with explanation please !!​

Answer 1

Answer:

option b is the correct answer.

Answer 2

Answer:

Glucose will boil at 100.104°C

Explanation:

Given that,

Molal elevation Constant for water is

$K _{b} = 0.52Km {}^{ - 1}$

Molality of glucose=0.2m

The Van'toff Factor for glucose is

$i = 1$

To find the boiling point,we use the relation

$∆T _{b}=k _{b} \times m \times i$

Substituting the known values,we get

$∆T _{b} = 0.52 \times 0.2 \times 1 = 0.104 {}^{0} C$

Therefore,the glucose will boil at

$T _{b} = 100 + 0.104 = 100.104 {}^{0} C$

Hence,glucose will boil at 100.104°C

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