Answer :

Given-

• molar mass of NaCl= 58.5 g mol^{-1}

• mass of water= 37.2g

• K_{f} for water = 1.86 K kg mol^{-1}

We know that, freezing point of water is 0^{0}C. By lower the freezing point by 2^{0}C we get the freezing point of Water to be -2^{0}C.

Depression of freezing poit for dilute solutions is directly proportional to molality m of the solution.

ΔT m

ΔT = K_{f} × m ………………….(1)

Where K_{f} 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= 0^{0}C-(-2^{0}C) = 2^{0}C

w_{2} grams of NaCl of Molar mass M_{2} is dissolved in w_{1} grams of water of molar mass M_{1}. This produces depression in freezing point of the solvent. Then molality of the solute is given by

Substituting m in equation (1) we get,

Rearranging,

Substituting given values in the above formula,

Weight of Nacl to produce depression in freezing point of water by 2^{0}C is 2.34g .

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