The correct relation between elevation of boiling point and molar mass of solute is

Question

The correct relation between elevation of boiling point and molar mass of solute is (A) M2 = (Kb . W2/Δ Tb.W1)  (B) M2 = (Kb . W1/Δ Tb.W2)  (C) M2 = (Tb . Kb/W1.W2)  (D) M2 = (Δ Tb . W1/Kb.W2)

Tardigrade Admin · Accepted Answer

For dilute solutions, the elevation in boiling point is directly proportional to the molal concentration of the solute in solution

(m)

Thus,

Δ T_{b} \propto m

Δ T_{b} = k_{b} m

Where,

k_{b}

is molal boiling point elevation constant.

As, molality

(m) = \frac{Moles of solute ( n _{2} )}{Mass of solvent ( W _{1} ) ( in k g )}

Moles of solute

(n_{2}) = \frac{W _{2}}{M _{2}}

∴ Δ T_{b} = \frac{k _{b} \times W _{2}}{M _{2} \times W _{1}}

\Rightarrow M_{2} = \frac{k _{b} \cdot W _{2}}{Δ T _{b} \cdot W _{1}}

Q. The correct relation between elevation of boiling point and molar mass of solute is

$M_{2} = \frac{K _{b} . W _{2}}{Δ T _{b} . W _{1}}$

$M_{2} = \frac{K _{b} . W _{1}}{Δ T _{b} . W _{2}}$

$M_{2} = \frac{T _{b} . K _{b}}{W _{1} . W _{2}}$

$M_{2} = \frac{Δ T _{b} . W _{1}}{K _{b} . W _{2}}$

Q. The correct relation between elevation of boiling point and molar mass of solute is

M2​=ΔTb​.W1​Kb​.W2​​

M2​=ΔTb​.W2​Kb​.W1​​

M2​=W1​.W2​Tb​.Kb​​

M2​=Kb​.W2​ΔTb​.W1​​

$M_{2} = \frac{K _{b} . W _{2}}{Δ T _{b} . W _{1}}$

$M_{2} = \frac{K _{b} . W _{1}}{Δ T _{b} . W _{2}}$

$M_{2} = \frac{T _{b} . K _{b}}{W _{1} . W _{2}}$

$M_{2} = \frac{Δ T _{b} . W _{1}}{K _{b} . W _{2}}$