### Example 3.17: Determining the Mass of Solute in a Given Volume of Solution

How many grams of $$\ce{NaCl}$$ are contained in 0.250 L of a 5.30 molar solution?

#### Solution

$$m_{\mathrm{NaCl}}$$ = ? ...(goal of this calculation)

$$c_{\mathrm{NaCl}} = 5.30\ \mathrm{M}$$

$$V_{\mathrm{solution}} = 0.250\ \mathrm{L}$$

The volume and concentration of the solution are specified, so the chemical amount (mol) of solute is easily computed as demonstrated in example 3.15:

$$n_{\mathrm{NaCl}} = c_{\mathrm{NaCl}} \cdot V_{\mathrm{solution}}$$

$$\ \ \ =5.30\ \mathrm{M} \cdot 0.250\ \mathrm{L}$$

$$\ \ \ =1.325\ \mathrm{mol}$$

Finally, this chemical amount is used to derive the mass of NaCl:

$$m_{\mathrm{NaCl}} = n_{\mathrm{NaCl}} \cdot M_{\mathrm{\ce{NaCl}}}$$

$$\ \ \ =1.325\ \mathrm{mol} \cdot 58.44\ \frac{\mathrm{g}}{\mathrm{mol}}$$

$$\ \ \ =77.4\ \mathrm{g}$$