Question

At 25°C, 100 g of water will be saturated with 35.7 g of NaCl. Which word below describes the solution of 1.55 mol of NaCl dissolved in 250 mL of water?

Answer 1

1) The saturation point at 25°C is 35.7 g of NaCl / 100 g of water => 35.7 %

Under normal circumstances water will not accept more salt than that.

2) The solution with 1.55 mol of NaCl dissolved in 250 mL of water =>

molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol

grams of NaCl = 1.55 mol * (molar mass of NaCl) = 1.55mol * 58.44 g/mol = 90.58 grams of NaCl


grams of water = 250 mL * 1 g/mL = 250 g/g.

Concentration of the solution:  [90.58 g NaCl / 250 g H2O] * 100 = 36.23 %

3) Conclusion: the solution has more salt than the saturation value. This means that the solution is supersaturated.

Supersaturation is a special condition, which is unstable, but that is not part of the questions.

The answer is supersaturated,