# Question be392

Mar 15, 2017

We need the quotient, $\text{moles of solute"/"volume of solution}$ to solve for $\text{normality}$ (i.e. $\text{molarity}$).

#### Explanation:

We need the quotient, $\text{moles of solute"/"volume of solution}$

In $100 \cdot g$ of SOLUTION there are $17 \cdot g$ of $H C l$ solute; the volume of this solution is "Mass"/("density"(rho))=(100*g)/(1.21*g*mL^-1) $=$ $82.6 \cdot m L$.

Given this we can calculation the concentration with respect to molarity:

$\text{Concentration} = \frac{\frac{17 \cdot g}{36.46 \cdot g \cdot m o {l}^{-} 1}}{82.6 \cdot m L \times {10}^{-} 3 \cdot m L \cdot {L}^{-} 1} \cong 6.0 \cdot m o l \cdot {L}^{-} 1$

The most concentrated hydrochloric acid you can buy("mass"/"mass")xx100% is approx. 36%. This corresponds to a concentration of approx. $12.0 \cdot m o l \cdot {L}^{-} 1$. Note that all I have done here is approach the problem dimensionally, and most of the problems you encounter as an undergrad may be solved by this method.

Also note that pure $H C l$ would be a room temperature gas ("boiling point"=-85.5 ""^@C)#.