# Question 54da7

Feb 1, 2018

$= 1.4283 M$

#### Explanation:

1. Assume the following data for the density of water and nitric acid.
$H N {O}_{3} = 1.42 \text{ g/ml}$ ("concentrated")
${H}_{2} O = 1.00 \text{ g/ml}$  ("at " 4^oC)
2. Since the provided data are in grams, convert it to volume. Observe uniformity of units; i.e.,
${V}_{H N {O}_{3}} = \frac{630 \cancel{g}}{\frac{1.42 \cancel{g}}{m l}} = 443.6620 m L \approx 443.7 m L$
${V}_{{H}_{2} O} = \frac{6500 \cancel{g}}{\frac{1.00 \cancel{g}}{m l}} = 6500 m L$
3. Find the volume of the solution. Note that volume solution is the summation of the volume of the solute and the solvent as shown below.
$\text{Volume"_("solution")="volume solute"+"volume solvent}$

"Volume"_("solution")=443.7mL+6500mL

"Volume"_("solution")=6943.7cancel(mL)xx(1L)/(1000cancel(mL))=6.9437L~~7.0000L#
4. Find the mole value of nitric acid $\left(\eta H N {O}_{3}\right)$ by taking the quotient of its $\text{mass} = 630 g$ against its $\text{molar mass"=63.012" g/mol}$; so that,
$\eta = \frac{m H N {O}_{3}}{M m H N {O}_{3}}$
$\eta = \frac{630 g}{\frac{63.012 g}{m o l}}$
$\eta = 9.998 m o l$
5. From the formula $\text{Molarity"(M)=("number of moles"(eta))/("Li solution} \left(V\right)$, find the concentration of the solution.
$M = \frac{\eta}{V}$
$M = \frac{9.998 m o l}{7.0000 L}$
$M = 1.4283 \text{ mol/L} \approx 1.4283 M$