Question

1) Boiling point of a solvent is 80.2°C. When 0.419
g of the solute of molar mass 252.4 g mol-' was
dissolved in 75 g of the solvent, the boiling point
of the solution was found to be 80.256°С. Find
the molal elevation constant. (2.53 K kg mol-!)​

Answer 1

Answer:Given: T(zero)0=(273+80.2)K

Tb=273+80.246K

W1= 75g

W2=0.419g

M2=252.4gmol-1

Kb=?

^Tb=Tb-To

=(273+80.256)-(273+80.2)

=0.056K

^Tb=Kb×W2×1000/W1×M2

Kb=^Tb×W1×M2/W2×1000

=0.056×75×252.4/0.41×1000

=2.53K kg mol-1

Explanation:

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Answer 2

Given:

• The boiling point of the solvent ($T_b$) = 80.2°C

• The boiling point of solution (T) = 80.256°C

• Weight of the solvent (w) = 75g

• Weight of solute  (W) = 0.419 g
• Molecular weight of solute (M) = 252.4 g

To Find:

• The molal elevation constant.

Solution:

First, we are going to find the elevation in boiling point which is given by,

⇒ Δ$T_b$ = $T-T_b$  → (equation 1)

Where 'T' is the boiling point of the solution and '$T_b$' is the boiling point of the solvent. Substitute the values from the given data in equation 1. We get,

⇒ Δ$T_b$ = 80.256-80.2 = 0.056K

When elevation in boiling point is proportional to molarity then,

Δ$T_b$ = $\frac{1000*K_b*W}{M*w}$  → (equation 2)

Where 'W' is the weight of solute, 'w' is the weight of solvent, 'M' is the molecular weight of solute, and '$K_b$' is the molal elevation constant.

Rearranging equation 2 to get the value of molal constant.

⇒ $K_b$ = $\frac{0.056*252.4*75}{0.419*1000}$ = 1060.08/419 = 2.53002 K Kg/mol

∴ $K_b$ = 2.53002 K Kg/mol.

∴ The molal constant = 2.53002 K Kg/mol.