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Standard XII

Chemistry

Depression in Freezing Point

Substance A...

Question

Substance $A$ tetramerises in water to the extent of 80%. A solution of $2.5 g$ of $A$ in $100 g$ of water lowers the freezing point by ${0.3}^{\circ} C$ . The molar mass of $A$ is:
( $K_{f}$ for water $=$ $1.86 K$ $k g$ $m o l^{- 1}$ )

122

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244

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Solution

The correct option is D 62
Given,

$i = \frac{0.8}{2} = 0.4$

Using the relation, $Δ T_{f} = i K_{f} m$

$m = \frac{Δ T_{f}}{i K_{f}}$

$\Rightarrow \frac{\frac{2.5}{M}}{0.1025} = \frac{0.3}{0.4 \times 1.86}$

$\Rightarrow M = 62 g / m o l$

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Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr₂ in 200 g of water (Molar mass of MgBr₂ is 184 g) (K_f for water is 1.86 K kg /mol)

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Substance A tetramerises in water to the extent of 80%. A solution of 2.5g of A in 100g of water lowers the freezing point by 0.3∘C. The molar mass of A is: (Kf for water = 1.86K kg mol−1)

The correct option is D 62Given,i=0.82=0.4Using the relation, ΔTf=iKfmm=ΔTfiKf⇒2.5M0.1025=0.30.4×1.86⇒M=62 g/mol

Substance $A$ tetramerises in water to the extent of 80%. A solution of $2.5 g$ of $A$ in $100 g$ of water lowers the freezing point by ${0.3}^{\circ} C$ . The molar mass of $A$ is:
( $K_{f}$ for water $=$ $1.86 K$ $k g$ $m o l^{- 1}$ )

The correct option is D 62
Given,

$i = \frac{0.8}{2} = 0.4$

Using the relation, $Δ T_{f} = i K_{f} m$

$m = \frac{Δ T_{f}}{i K_{f}}$

$\Rightarrow \frac{\frac{2.5}{M}}{0.1025} = \frac{0.3}{0.4 \times 1.86}$

$\Rightarrow M = 62 g / m o l$