How many grams of urea should be dissolved in 500 gram water so that the depression in freezing point is 0.2 K? Kf for solvent is 3.2 K kg mole⁻¹.Solve the example.
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Answer:
1.875 gm.
Explanation:
Quantity of water = 500gm (Given)
Freezing point = 0.2K (Given)
Kf solvent = 3.2 (Given)
Depression in Freezing point -
ΔT = i × kf × m
where m is the molarity of the solvent , Kf is the cryoscopic constant and t is the temperature.
Therefore,
0.2 = 3.2 × x/ 60 × 0.5
x = 30 × 2/32
x = 30/16
x = 1.875
Thus, the amount of uread required is 1.875 gm.
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