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Standard XII

Chemistry

Elevation in Boiling Point

A solution co...

Question

A solution containing $12.5 g$ of non-electrolyte substance in $185 g$ of water shows boiling point elevation of $0.80 K$ . Calculate the molar mass of the substance. ( $K_{b} = 0.52 K$ $k g$ ${m o l}^{- 1}$ )

53.06 g

{m o l}^{- 1}

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25.3 g

{m o l}^{- 1}

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16.08 g

{m o l}^{- 1}

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43.92 g

{m o l}^{- 1}

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Solution

The correct option is B $43.92 g$ ${m o l}^{- 1}$
${Δ T}_{b} = K_{b} \times$ molality
$0.80 = 0.52 \times \frac{12.5 \times 1000}{M_{w} \times 185}$
$M_{w} = \frac{0.52 \times 12.5 \times 1000}{185 \times 0.8}$
$M_{w} = 43.92$ $g$ ${m o l}^{- 1}$

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A solution containing 12.5g of non-electrolyte substance in 185g of water shows boiling point elevation of 0.80K. Calculate the molar mass of the substance. (Kb=0.52K kg mol−1)

The correct option is B 43.92g mol−1ΔTb=Kb× molality0.80=0.52×12.5×1000Mw×185Mw=0.52×12.5×1000185×0.8Mw=43.92 g mol−1

A solution containing $12.5 g$ of non-electrolyte substance in $185 g$ of water shows boiling point elevation of $0.80 K$ . Calculate the molar mass of the substance. ( $K_{b} = 0.52 K$ $k g$ ${m o l}^{- 1}$ )

The correct option is B $43.92 g$ ${m o l}^{- 1}$
${Δ T}_{b} = K_{b} \times$ molality
$0.80 = 0.52 \times \frac{12.5 \times 1000}{M_{w} \times 185}$
$M_{w} = \frac{0.52 \times 12.5 \times 1000}{185 \times 0.8}$
$M_{w} = 43.92$ $g$ ${m o l}^{- 1}$