हिंदी

The freezing point of a solution containing 0.3 gms of acetic acid in 30 gms of benzene is lowered by 0.45 K. Calculate the Van’t Hoff factor. (at. wt. of C = 12, H = 1, O = 16, Kf for benzene = 5.12 - Chemistry (Theory)

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प्रश्न

The freezing point of a solution containing 0.3 gms of acetic acid in 30 gms of benzene is lowered by 0.45 K. Calculate the Van’t Hoff factor. (at. wt. of C = 12, H = 1, O = 16, Kf for benzene = 5.12 K kg mole−1). 

The freezing point of a solution containing 0.3 g of acetic acid in 30.0 g of benzene is lowered by 0.45°C. Calculate the van’t Hoff factor. (Kf for benzene = 5.12 K kg mol−1)

योग
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उत्तर

Given: ΔTf = 0.45

w = 0.3 g

W = 30g

Mol. mass (M) of acetic acid, CH3COOH = 60 

K= 5.12

If i is the Van’t Hoff factor, then depression in freezing point

`Delta T_f = (i xx 1000 xx K_f xx w)/(W xx M)`

`i = (Delta T_f  xx W xx M)/(1000 xx K_f xx w)`

i = `(0.45 xx 30 xx 60)/(1000 xx 5.12 xx 0.3)`

i = `810/1536`

i = 0.5273

∴ Van’t Hoff factor is 0.5273.

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Relative Molecular Mass - Freezing Point Depression
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नूतन Chemistry Part 1 and 2 [English] Class 12 ISC
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REVIEW EXERCISES | Q 2.103 | पृष्ठ १०५
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