Calculate the mass of Nacl (molar mass = 58.5 g mol^-1) to be dissolved in 37.2g of water to lower the freezing point by 2⁰C, assuming that NaCl undergoes complete dissociation. (K_f for water = 1.86 K kg mol^-1).

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 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 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= 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,

Rearranging,

Substituting given values in the above formula,

Weight of Nacl to produce depression in freezing point of water by 2⁰C is 2.34g .