Question

SHOW ALL WORK AND STEPS NO HAND WRITTEN WORK

B.) calculate molarity of the solution

2. An unlabeled bottle contains an unknown potassium halide salt, KX. A student must determine the identity of the salt of KX yields dissolved in 24.65 g of distilled water, the freezing point of the resulting solution is-1.01 °C. The freezing point of the distilled water is determined to be +0.02°C. by using freezing point depression data. Because KX is a strong electrolyte, one mole two moles of ions in solution. The student finds that when 1.13 g of the unknown salt is

Answer 1

Let the molar mass of solute = M.

Now molality = (1.13/M)/0.02465

Change in temperature = 0.02-(-1.01) = 1.03C

So, we know deltaT = Kf*m

1.03 = 1.86 x (1.14/M)/0.02465

M = 83.52 g/mol

Moles of solute = 1.13/83.52= 0.01353 mol

Volume = 24.65 g of distilled water = 24.65ml of water = 0.02465 L distilled water

So, molarity = 0.01353/0.02465 = 0.5489 M

If you satisfied with the solution please like it thankxx...