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Elevation in Boiling Point

117 g of NaCl...

Question

$117 g$ of $N a C l$ is added to $222 g$ of water in a container. Calculate the temperature at which the water boils at 1 atm?
Given:
$Ebullioscopic constant for water (K_{b}) = 0.52 K k g {mol}^{- 1}$
$Molecular mass of NaCl = 58.5 g m o l^{- 1}$
and boiling point of pure water is $100^{\circ} C$

{98.3}^{\circ} C

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{102.8}^{\circ} C

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{104.7}^{\circ} C

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{101.5}^{\circ} C

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Solution

The correct option is C ${104.7}^{\circ} C$
$Weight of solvent w_{1} = 222 g$
$Weight of solute w_{2} = 117 g$
$M_{2} = Molecular mass of NaCl = 58.5 g m o l^{- 1}$
$K_{b} = 0.53 K k g {mol}^{- 1}$

Now, addition of a non-volatile solute causes elevation in boiling point i.e. $Δ T_{b}$
$Δ T_{b} = \frac{(k_{b} \times 1000 \times w_{2})}{(M_{2} \times w_{1})}$
On substituting,
$Δ T_{b} = \frac{(0.52 \times 1000 \times 117)}{(58.5 \times 222)} Δ T_{b} = {4.7}^{\circ} C T_{b} - 100^{\circ} C = {4.7}^{\circ} C$
$T_{b} = (100 + 4.7)^{\circ} C = {104.7}^{\circ} C$ .

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Elevation in Boiling Point

Standard XII Chemistry

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117 g of NaCl is added to 222 g of water in a container. Calculate the temperature at which the water boils at 1 atm? Given: Ebullioscopic constant for water(Kb)=0.52 K kg mol−1 Molecular mass of NaCl=58.5 g mol−1 and boiling point of pure water is 100∘C

$117 g$ of $N a C l$ is added to $222 g$ of water in a container. Calculate the temperature at which the water boils at 1 atm?
Given:
$Ebullioscopic constant for water (K_{b}) = 0.52 K k g {mol}^{- 1}$
$Molecular mass of NaCl = 58.5 g m o l^{- 1}$
and boiling point of pure water is $100^{\circ} C$