The boiling point of a solution of 0-512 gm of napthalene- C10H8- in 24 gm of chloroform is 0-5-C higher than that of pure chloroform- The molal boiling point elevation constant of chloroform is in K-kg-mol

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Standard VIII

Chemistry

Boiling

The boiling p...

Question

The boiling point of a solution of $0.512 gm$ of napthalene, $C_{10} H_{8},$ in $24 gm$ of chloroform is ${0.5}^{\circ} C$ higher than that of pure chloroform. The molal boiling point elevation constant of chloroform is (in K-kg/mol)

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Solution

Boiling point elevation constant
$Δ T_{b} = K_{b} \times m . . . . . (1)$
Where $Δ T_{b} =$ change in boiling point
$K_{b} =$ boiling point elevation constant
$m$ = molality
Molality $(m) = \frac{mole of solute}{mass of solvent in kg}$
$\Rightarrow \frac{0.512}{128 \times 24} \times 1000 = 0.166$
Given that
$Δ T_{b} = {0.5}^{\circ} C$
Put all the values in equation (1), we get
$\Rightarrow 0.5 = 0.166 \times K_{b}$
$\Rightarrow K_{b} = 3$

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The boiling point of a solution of 0.512 gm of napthalene, C10H8, in 24 gm of chloroform is 0.5∘C higher than that of pure chloroform. The molal boiling point elevation constant of chloroform is (in K-kg/mol)

The boiling point of a solution of $0.512 gm$ of napthalene, $C_{10} H_{8},$ in $24 gm$ of chloroform is ${0.5}^{\circ} C$ higher than that of pure chloroform. The molal boiling point elevation constant of chloroform is (in K-kg/mol)