# How many grams of calcium nitrate are needed to make 3.30L of a 0.10 M solution?

Apr 3, 2018

See below

#### Explanation:

We'll use "Molarity" = ("moles")/("liter"), and rearrange to get $\text{moles" = "Molarity" xx "Liters}$

$0.1 M \times 3.30 L = 0.33 \text{moles}$

Molar mass of Calcium Nitrate is 164.1g/mol

$0.33 \text{moles" xx 164.1g/"mol}$ = 54.15 g $C a {\left(N {O}_{3}\right)}_{2}$