D. CuCl2 copper(2)chloride
Please provide whatever follows your question. Without it we cannot answer the question.
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,
*brings you good news*
The answer is D. 12
Hope i will help you angain sometime :)
~~~~~SoraKeyblader~~~~~
