If the elevation in boiling point of a solution of 10 g of solute ( mol. wt. = 100 ) in 100 g of water is $Δ T_{b}$ , the ebullioscopic constant of water is:

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100 T_{b}

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Δ T_{b}

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\frac{Δ T_{b}}{10}

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Solution

The correct option is C $Δ T_{b}$
We know that,

$Δ T_{b} = \frac{1000 \times K_{b} \times w}{W \times M}$

Substituting the values,

$Δ T_{b} = \frac{1000 \times K_{b} \times w}{W \times M}$

$Δ T_{b} = \frac{1000 \times K_{b} \times 10}{100 \times 100}$

$Δ T_{b} = K_{b}$

Option C is correct.

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