If the solubility of KCl in water at 95° C is 55 g/100 g. Express the solubility in molarity.

Answer 1

I assume that is 100 g OF THE SOLUTION. SO, what's the density of the solution? You will need to know that. I will assume the density of the solution is 1.05 g/mL but you should use the value you find. It will NOT be 1.00 g/mL.
Density of the solution is 1.05 g/mL; therefore, mass = volume x density and 100 g = volume x 1.05 g/mL so volume = 100/1.05 = 95.2 mL therefore you have 55 g KCl in 95.2 mL solution.
M = mols/L. and mols = g/molar mass with volume = 0.952 L. I'll let you finish. Post your work if you get stuck.

Answer 2

Hello!
So I didn’t get a a density! Here are the questions I was given and I’m trying to do part c,

A. (3 pts) To the nearest whole number, what is the solubility of KC1 in g/100 g of water at 95
°C? Hint you must interpret the scale correctly for full credit.

I put this answer
55.0 g/100 g
B. (4 pts) Express the solubility of KC1 in water at 95 °C (from part a) in units of grams per
liter. Remember that 1 g = 1 mL = 1 cm? for water. Ignore any displacement effects.

I put this answer
0.055 g/1 L
C. (5 pts) Express the solubility of KC1 in water at 95 °C (from part a) in molarity (M). Hint-
requires molar mass.

I have this down already:
55 g KC1 / 100 g H2) x1 g / 1 mL x 1000 mL / L x (39.0983 + 35.453) =
55 g KCl / 100 g H2) x 1 g / 1 mL x 1000 mL / L x (74.5513) =