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Relative Lowering of Vapour Pressure

How many gram...

Question

How many grams of sucrose (Molecular weight $= 342$ ) should be dissolved in $100 g m$ water in order to have a solution with sum of lowering of freezing point and elevation of boiling point equal to $10^{0} C$ $(K_{b} = 0.51, K_{f} = 1.86)$ ?

34.2 g m

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72.15 g m

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342 g m

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144.3 g m

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Solution

The correct option is C $144.3 g m$
$△ T_{f} = K_{f} \cdot m, △ T_{b} = K_{b} \cdot m$
$△ T_{f} + △ T_{b} = 10 = (K_{f} + K_{b}) \cdot m$
$∴ m = \frac{10}{0.51 + 1.86} = 4.22$
$∴ \frac{20 \times 1000}{342 \times 100} = 4.22$
$W = 144.3 g m$

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How many grams of sucrose (Molecular weight=342) should be dissolved in 100gm water in order to have a solution with sum of lowering of freezing point and elevation of boiling point equal to 100C (Kb=0.51,Kf=1.86) ?

The correct option is C 144.3gm△Tf=Kf⋅m,△Tb=Kb⋅m△Tf+△Tb=10=(Kf+Kb)⋅m∴m=100.51+1.86=4.22∴20×1000342×100=4.22W=144.3gm

How many grams of sucrose (Molecular weight $= 342$ ) should be dissolved in $100 g m$ water in order to have a solution with sum of lowering of freezing point and elevation of boiling point equal to $10^{0} C$ $(K_{b} = 0.51, K_{f} = 1.86)$ ?

The correct option is C $144.3 g m$
$△ T_{f} = K_{f} \cdot m, △ T_{b} = K_{b} \cdot m$
$△ T_{f} + △ T_{b} = 10 = (K_{f} + K_{b}) \cdot m$
$∴ m = \frac{10}{0.51 + 1.86} = 4.22$
$∴ \frac{20 \times 1000}{342 \times 100} = 4.22$
$W = 144.3 g m$