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प्रश्न
The freezing point of a solution containing 0·3gms of acetic acid in 30gms of benzene is lowered by 0.45K. Calculate the Van't Hoff factor. (at. wt. of C = 12, H = 1, O = 16, Kf for benzene = 5· 12K kg mole-1).
उत्तर
If i is the Van't Hoff factor, then depression in freezing point
`Delta T_f = (ixx 1000 xxK_f xx w)/(W xxM )`
Given, `Delta T_f = 0.45K,`
w = 0.3 g
W = 30g
Mol. mass (M) of acetic acid, CH3 COOH = 60
Kf = 5.12
`i = (Delta T_f xxWxxM)/(1000xxK_fxxw) = (0.45 xx 30xx60)/(1000 xx 5.12 xx 0.3)`
i = 0.5273
Van't Hoff factor is 0.5273.
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संबंधित प्रश्न
The molal freezing point constant of water is 1·86 K kg mol-1. Therefore, the freezing point of 0·1 M NaCl solution in water is expected to be:
1) -1.86°C
2) -0.372 °C
3) -0.186 °C
4) +0.372 °C
Define cryoscopic constant.
The freezing point of a solution containing 5.85 g of NaCl in 100 g of water is -3.348°C. Calculate Van’t Hoff factor ‘i’ for this solution. What will be the experimental molecular weight of NaCl?
(Kf for water = 1.86 K kg mol-1, at. wt. Na = 23, Cl = 35.5)
When 2 g of benzoic acid is dissolved in 25 g of benzene, it shows a depression in freezing point equal to 1.62 K. The molal depression constant (Kf) of benzene is 4.9 K kg mol-1, and the molecular weight of benzoic acid is 122 g/mol. What will be the percentage association of the benzoic acid?
(Benzoic acid forms dimers when dissolved in benzene.)
