What volume of a 0.771 M solution of CaCl2 contains 1.28 g of solute?

Answer 1

First we need to determine the molecular mass of CaCl2 which is 110.99 amu or 110.99 grams/mole.

A 0.771 M solution has 0.771 moles of CaCl2 per 1000 mL, so we can write 1000 mL / 0.771 moles = 1297 mL / mole.

Multiply 1297 mL/mole x 1 / 110.99 g/mole x 1.28 g CaCl2 = 15.0 mL.

Note that all units cancel except mL.

Always check the math.