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Freezing point of an aqueous solution is $\left ( -0.186 \right )^{\circ}C$ Elevation of boiling point of the same solution is $K_{b}= 0.512^{\circ}C,K_{f}= 1.86^{\circ}C$ find the increase in boiling point.

Option 1)
$0.186^{\circ}C$

Option 2)
$0.0512^{\circ}C$

Option 3)
$0.092^{\circ}C$

Option 4)
$0.2372^{\circ}C$

Answers (1)

best_answer

As we learnt in

Mathematical Expression -

$\Delta T_{b}= K_{b}\: m$

Unis of $K_{b}=\frac{K-K_{g}}{mole}$

$\Delta T_{b}= Elevation\: in \: boiling\: point$

- wherein

$K_{b}= Boiling \: point \: elevation \: constant$

$m= molality$

Mathematical Expression of Depression in Freezing point -

$\Delta T_{f}= K_{f}\: m$

- wherein

m = molarity of solvent

$K_{f}$ = cryoscopic constant

molal depress const

Units = $\frac{K-K_{g}}{mole}$

$\frac{T_{b}}{\Delta T_{f}}=\frac{K_{b}}{K_{f}}$

$\frac{\Delta T_{b}}{0.186}=\frac{0.512}{1.86}$

$\Delta T_{b}=0.0512 c^{o}$

Correct option is 2.

Option 1)

$0.186^{\circ}C$

This is an incorrect option.

Option 2)

$0.0512^{\circ}C$

This is the correct option.

Option 3)

$0.092^{\circ}C$

This is an incorrect option.

Option 4)

$0.2372^{\circ}C$

This is an incorrect option.

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