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प्रश्न
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)
उत्तर
ΔTf = 3.348°C
WB = 5.85g
WA = 100g
Kf = 1.86 K kg mol-1
Observed mol. mass of NaCl(MB)
`triangle "T"_"f" =" K"_"f" . ("W"_"B" xx 1000)/("W"_"A"xx"M"_"B")`
`triangle "M"_"B" = ("K"_"f" . "W"_"B"xx1000)/("W"_"A"xxtriangle "T"_"f")`
`= (1.86xx5.85xx 1000)/(100xx3.348)`
`= (18.6xx5.85)/3.348`
Observed mol. mass of NaCl = 32.5 g mol-1
Normal mol. mass of NaCl = 23 + 35.5 = 58.5 g mol-1
Van’t Hoff factor, i =`58.5/32.5`= 1.8
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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 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).
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.)
