Elevation of boiling point of 1 molar aqueous glucose solution (density of solvent = 1.2g/ml) is _________.

K_{b}

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1.20 K_{b}

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1.30 K_{b}

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0.833 K_{b}

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Solution

The correct option is A $K_{b}$
$D e n s i t y = 1.2 g / m l = 1.2 k g / L$

The concentration is 1 molar.

1 L of solution contains 1 mole of glucose.

$M a s s = D e n s i t y \times V o l u m e = 1.2 k g / L \times 1 L = 1.2 k g$

Thus 1.2 kg of solution contains 1 mole of glucose

mass of solvent = mass of solution - the mass of 1mole glucose

mass of solvent = $1200 g - 180 g$

mass of solvent = $1020 g = 1.02 k g$ .

$M o l a l i t y = \frac{number of moles of solute}{mass of solvent in kg} = \frac{1}{1.02} = 0.98$

$Δ T_{b} = m \times K_{b} = 0.98 k_{b} \approx 1 K_{b}$

Hence, the correct option is $A$

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