The rise in the boiling point of a solution containing 1.8g of glucose in 100g of solvent is ${0.1}^{o} C$ . The molal elevation constant of the liquid is:

{0.01}^{o} C m^{- 1}

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{0.1}^{o} C m^{- 1}

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1^{o} C m^{- 1}

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10^{o} C m^{- 1}

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Solution

The correct option is C $1^{o} C m^{- 1}$
Using formula,

$Δ T_{b} = K_{b} \times molality$
$⟹ 0.1 = K_{b} \times \frac{1.8 \times 1000}{180 \times 100}$
$⟹ 0.1 = K_{b} \times 0.1$

$⟹ K_{b} = 1^{o} C m^{- 1}$

Hence, the correct option is $C$

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The rise in the boiling point of a solution containing 1.8g of glucose in 100g of solvent is 0.1oC. The molal elevation constant of the liquid is:

The correct option is C 1oCm−1Using formula,ΔTb=Kb×molality⟹0.1=Kb×1.8×1000180×100⟹0.1=Kb×0.1⟹Kb=1 oC m−1Hence, the correct option is C

The rise in the boiling point of a solution containing 1.8g of glucose in 100g of solvent is ${0.1}^{o} C$ . The molal elevation constant of the liquid is:

The correct option is C $1^{o} C m^{- 1}$
Using formula,

$Δ T_{b} = K_{b} \times molality$
$⟹ 0.1 = K_{b} \times \frac{1.8 \times 1000}{180 \times 100}$
$⟹ 0.1 = K_{b} \times 0.1$

$⟹ K_{b} = 1^{o} C m^{- 1}$

Hence, the correct option is $C$