A student is instructed to make 0.250 liter of a 0.200 m aqueous solution of ca (no3) 2. what is a correct numerical setup for calculating the total number of moles of ca (no3) 2 needed to make 0.250 liter of the 0.200 m calcium nitrate solution?

The concentration is expressed in molarity which is equal to moles of solute per liters of solution. We use this definition to calculate the amount we need of the solute. We do as follows:

0.200 mol ca (no3) 2 / L solution (0.250 L solution) = 0.05 mol ca (no3) 2

0.05 mol ca (no3) 2 (164.09 g / mol) = 8.20 g ca (no3) 2 needed