Calculate the amount of sodium chloride (in g) which must be added to 1000 mL of water so that its freezing point is depressed by 0.744 K. For water, Kf = 1.86 K/m. Assume density of water to be 1 g mL-1.

Question

Calculate the amount of sodium chloride (in g) which must be added to 1000 mL of water so that its freezing point is depressed by 0.744 K. For water, Kf = 1.86 K/m. Assume density of water to be 1 g mL-1. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Mass of NaCl, w2 = ?, Volume of water = 1000 mL, Δ Tf = 0.744 K, Density of water = 1 g mL-1 So, mass of water W1 = 1000 mL × 1 g mL-1 = 1000g= 1 kg Δ Tf = iKf m =iKf ×((W258.5)/W1) = (iKf W2/58.5 × W1) W2 =(Δ Tf ×58.5×1/i × Kf) = (0.744 ×58.5/2×1.86) = 11.7 g (∵ i for NaCl =2) So, Mass of NaCl required = 11,7g.

Q. Calculate the amount of sodium chloride (in g) which must be added to 1000mL of water so that its freezing point is depressed by 0.744K. For water, Kf​=1.86K/m. Assume density of water to be 1gmL−1.

Q. Calculate the amount of sodium chloride (in g) which must be added to $1000 m L$ of water so that its freezing point is depressed by $0.744 K$ . For water, $K_{f} = 1.86 K / m$ . Assume density of water to be $1 g m L^{- 1}$ .