# Question #b082f

Jun 24, 2017

You must add 80 g of $\text{CaCl"_2·6"H"_2"O}$.

#### Explanation:

Step 1. Calculate the masses of ${\text{Na"_2"CO}}_{3}$ and of water in the solution

$\text{Mass of soln" = 47 color(red)(cancel(color(black)("mL solution"))) × "1.08 g solution"/(1 color(red)(cancel(color(black)("mL solution")))) = "50.8 g solution}$

${\text{Mass of Na"_2"CO"_3 = 50.8 color(red)(cancel(color(black)("g solution"))) × ("25 g Na"_2"CO"_3)/(100 color(red)(cancel(color(black)("g solution")))) = "12.7 g Na"_2"CO}}_{3}$

Step 2 Calculate the mass of $\text{CaCl"_2·6"H"_2"O}$ to be added

Let $x = \text{mass of CaCl"_2·6"H"_2"O}$. Then

$\text{Total mass" = (47 + x)color(white)(l)"g}$

$\frac{12.7}{47 + x} = 0.10$

$12.7 = 0.10 \left(47 + x\right) = 4.7 + 0.10 x$

$0.10 x = 8.0$

$x = \frac{8.0}{0.10} = \text{80 g}$

You must add 80 g of $\text{CaCl"_2·6"H"_2"O}$ to the solution.