Question

The freezing point of a 0.01 M aqueous solution of Nacl will be‚

Answer 1

Concept Introduction: Freezing point of a solution is the point at which it turns into solid.

Given:

We have been Given:

1. Molarity of aqueous NaCl solution is
2. $0.01m$

To Find:

We have to Find: The freezing point of the solution (∆Tf)

Solution:

According to the problem, Molarity is

$0.01m$

therefore, ∆Tf = Kf ×M.

Now, Kf of Water is,

$1.86k \times \frac{kg}{mole}$

therefore, according to the formula,

∆Tf is,

$1.86 \times 0.01 \times 2 = 1.86 \times 0.02 = 0.372k$

Hence, the Freezing point of aq. NaCl solution is

$0.372k$

Final Answer: The Freezing point of aq. NaCl solution is,

$0.372k$

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