# Calculate the molar concentration of the solution in 2 liters of which contained 58 g of lead(II) nitrate?

Apr 12, 2018

The concentration of the solution will be $\text{0.09 M}$.

#### Explanation:

The molar mass of $\text{Pb"("NO"_3)_2: " 332.2 g/mol}$

So

"58 g Pb"("NO"_3)_2 * "1 mol"/("331.2 g Pb"("NO"_3)_2) = "0.0175 mol of Pb"("NO"_3)_2

The formula for the molar concentration is

$\text{M"="mol"/"L}$

Since $\text{L}$ is given and the mol is calculated,

$\text{M" = "0.0175 mol"/"2 L}$

The answer will be around $\text{0.0875 M}$. However, you need only one sigfig, so you round it, which gives you $\text{0.09 M}$.