Question

The boiling point of an aqueous solution of a non volatile solute is 100.15 what is the freezing pointquora

Answer 1

The freezing point is -0.272 degree Celsius is the freezing point.

Explanation:

• For finding the boiling point we will use the van’t Hoff factor which is

        $T_b = i \times K_b\times m.$

$K_b$ and $K_f$ for water are 0.512 and 1.86 K $molality^-1.$

• ‘i’ is given as 1 because it is a non-volatile solute.

• Therefore $0.15 = 1 \times 0.512 \times m.$

• This gives us the value of m which is equal to 0.2929.

• Now when we add the volume of the water we get $m \times v -= m' \times 2v.$

• Therefore m' is equal to $\frac{0.2929}{2}.$  which gives us a value of 0.1464.

• Therefore the temperature of freeing is $1 \times$ $K_f \times m'$ which is $1 \times 0.1464 \times1.86$ which is equivalent to 0.272.

• Therefore to find the $T_f$ we need to apply the formula of

$T_f - Tf_o =0- 0.272$ which gives us a value of -0.272 degree Celsius.

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