Question

In 250 mL of solution the following is dissolved:

40 mg calcium chloride

40 mg calcium citrate

800 mg sodium chloride

75 mg potassium chloride

20 mg sodium lactate

Find the osmolarity of the solution.

Answer 1

no of mole of CaCl2 = 40/111 = 0.36 mmole

no of mole of calcium citrate = 40/498.5 = 0.08024 mmole

no of mole of sodium chloride = 800/58.5 = 13.67 mmole

no of mole of sodium lactate = 75/112.06 = 0.67 mmole

total no of moles = 0.36+0.08024+13.67+0.67 = 14.78 mmole

osmolarity = no of mole of solutes / volume of solution

           = 14.78 /250

           = 0.059 M