Question

What is the molarity of a solution consisting of 105 grams of calcium nitrate and 280. grams of water?

Answer 1

About 2.29 M.

Step-by-step explanation:

Recall that molarity is defined by moles of solute over liters of solution.

Our solute is 105 grams of calcium nitrate (Ca(NO₃)₂)).

Convert 105 grams of Ca(NO₃)₂ to moles. The molecular weight of Ca(NO₃)₂ is 164.10 g/mol:

`\displaystyle 105\text{ g Ca(NO$_3$)$_2$}\cdot \frac{1\text{ mol Ca(NO$_3$)$_2$}}{164.10\text{ g Ca(NO$_3$)$_2$}} = 0.640\text{ mol Ca(NO$_3$)$_2$}}`

Convert 280. grams of water to liters. Recall that the density of water is given by 1.00 g/mL:

`\displaystyle \begin{aligned} D & = (m)/(V) \\ \\ V & = (m)/(D) \\ \\ & = \frac{(280.\text{ g})}{\left(\frac{1.00\text{ g}}{1\text{ mL}}\right)} \\ \\ & = 280. \text{ mL}\cdot \frac{1\text{ L}}{1000\text{ mL}} = 0.280\text{ L soln.}\end{aligned}`

Hence, the molarity of the solution is:

`\displaystyle \begin{aligned} \text{M} &= \frac{\text{ mols solute}}{\text{L soln.}} \\ \\ & = \frac{(0.640\text{ mol})}{(0.280\text{ L soln.})} \\ \\ & = 2.29\text{ M} \end{aligned}`

In conclusion, the solution is about 2.29 M.