• molar mass of NaCl= 58.5 g mol-1
• mass of water= 37.2g
• Kf for water = 1.86 K kg mol-1
We know that, freezing point of water is 00C. By lower the freezing point by 20C we get the freezing point of Water to be -20C.
Depression of freezing poit for dilute solutions is directly proportional to molality m of the solution.
ΔT = Kf × m ………………….(1)
Where Kf freezing point depression constant
ΔT is obtained by subtracting freezing point of solution from freezing point of pure solvent i.e.,
In this problem, ΔT= 00C-(-20C) = 20C
w2 grams of NaCl of Molar mass M2 is dissolved in w1 grams of water of molar mass M1. This produces depression in freezing point of the solvent. Then molality of the solute is given by
Substituting m in equation (1) we get,
Substituting given values in the above formula,
Weight of Nacl to produce depression in freezing point of water by 20C is 2.34g .
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