How many moles HCl are contained in $3 L$ of a nitric acid solution at 12.33% ($\rho$ $ =$ $1.070\frac{g}{cm^3}$) ?
Let's refer to the $12.33%$ nitric acid solution by $S$: $V_S$ $=$ $3000 cm^3$ 1)$m_S$ $=$ $V_S \cdot \rho_S$ $ = $ $3000 \cdot 1.070$ $ = $ $3210g$ (Units!) 2)$m_{HNO_3}$ $=$ $\frac{%_S \cdot m_S}{100}$ $=$ $\frac{12.33 \cdot 3210}{100}$ $=$ $395.8g$ 3)$n_{HNO_3}$ $=$ $\frac{m_{HNO_3}}{M_{HNO_3}}$ $=$ $\frac{395.8}{63}$ $=$ $6.28 mol$
