Addition of 0-5 g of a compound to 50 mL of benzene having density -0-9 g mL-1 lowers the freezing point from 5-51-C to 5-01-C- Molar mass of the compound added is-Kf of benzene - 5-12 K kg mol-1-

Weight (w_2) of benzene = V_1 × ρ nbsp;Number of moles of compound(solute) added n= ( w_1m_1 ) On adding n moles of solute to w_2 g of solvent, depression ...

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Standard XII

Chemistry

Depression in Freezing Point

Addition of 0...

Question

Addition of $0.5 g$ of a compound to $50 m L$ of benzene $( having density ρ \approx 0.9 g m L^{- 1})$ lowers the freezing point from ${5.51}^{\circ} C$ to ${5.01}^{\circ} C$ . Molar mass of the compound added is:
$(K_{f}$ of benzene = $5.12 K k g m o l^{- 1}$ ).

185.17 g m o l^{- 1}

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221.75 g m o l^{- 1}

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55.61 g m o l^{- 1}

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113.78 g m o l^{- 1}

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Solution

The correct option is D $113.78 g m o l^{- 1}$
weight $(w_{2})$ of benzene = $V_{1} \times ρ$

$Number of moles of compound(solute) added n = (\frac{w_{1}}{m_{1}})$

On adding $n$ moles of solute to $w_{2} g$ of solvent, depression in freezing point is:

$Δ T_{f} = \frac{1000 \times K_{f} \times w_{1}}{m_{1} \times w_{2}}$

$(5.51 - 5.01) = \frac{1000 \times 5.12 \times 0.5}{m_{1} \times (50 \times 0.9)}$

$∵ m_{1} = 113.78 g m o l^{- 1}$

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Addition of 0.5 g of a compound to 50 mL of benzene ( having density ρ≈0.9 g mL−1) lowers the freezing point from 5.51∘C to 5.01∘C. Molar mass of the compound added is: (Kf of benzene = 5.12 K kg mol−1).

The correct option is D 113.78gmol−1weight (w2) of benzene = V1 × ρ Number of moles of compound(solute) added n=(w1m1) On adding n moles of solute to w2 g of solvent, depression in freezing point is: ΔTf=1000×Kf×w1m1×w2 (5.51−5.01)=1000×5.12×0.5m1×(50×0.9) ∵m1=113.78gmol−1

Addition of $0.5 g$ of a compound to $50 m L$ of benzene $( having density ρ \approx 0.9 g m L^{- 1})$ lowers the freezing point from ${5.51}^{\circ} C$ to ${5.01}^{\circ} C$ . Molar mass of the compound added is:
$(K_{f}$ of benzene = $5.12 K k g m o l^{- 1}$ ).