Question

The boiling point of 0.2M solution of cane sugar is (Kb)=0.52K/m
[A]100.2°C
[B]100.104°C
[C]100.313°C
[D]101.1°C​

Answer 1

Answer:

c is correct

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Answer 2

Given,

Cane sugar ($C_{12}H_{22}O_{11}$) with

Molarity, $m=0.2M$

Molar boiling point constant, $K_{b}=0.52K/m$

To find,

The boiling point of cane sugar, $b.p$

Solution,

We know that,

To find boiling point we need to add boiling point elevation to the standard boiling point of water.

⇒$b.p= 100+$Δ$T_{f}$

To find Δ$T_{f}$

Δ$T_{f}=K_{b}$ × $m$ × $i$

Where $i=$ Van Hoff's factor

Now,

Van Hoff's factor for $C_{12}H_{22}O_{11}$ is $1$ because it is an organic substance hence shows no dissociation.

So,

Δ$T_{f}=0.52$ × $0.2$ × $1$

⇒Δ$T_{f}=0.104$

And hence,

Boiling point of cane sugar is

$b.p=100+0.104$

⇒$b.p=100.104$°C

Therefore, the boiling point of cane sugar is $100.104$°C. (Option B)