# Question 06b85

Nov 18, 2015

19.99 g CaCl

#### Explanation:

This is a dilution problem. The equation for dilution is:
$\left({M}_{1}\right) \left({V}_{1}\right) = \left({M}_{2}\right) \left({V}_{1}\right)$

You're given the mass of calcium chloride and the volume of solution. The first step is to find the amount of moles of CaCl that you have. Using dimensional analysis you can solve.

$40 g C a C l$$\times$$\frac{1 m o l}{75.528 g C a C l}$=$.5296 m o l$

Now we find the molarity by dividing the moles of CaCl by the volume of solution given in the question. Convert the solution into liters, as it's the units you will need to finish the problem.

$\frac{.5296 m o l C a C}{.65 l}$=$.81470 M$

This is the ${M}_{1}$ in the dilution equation. So plugging in the numbers into the equation it looks like this. 2 L is give to us in the problem and since it's in liters we don't have to convert.

$\left(0.8147\right) \left(.65\right)$=$\left({M}_{2}\right) \left(2\right)$

Rearranging using some algebra we get

$\left(0.8147\right) \left(.65\right)$/$\left(2\right)$ = $0.2647 m o l$

Now that we have moles of CaCl we can convert into required mass.

$0.2647$ $m o l$ $C a C l$ x (75.528gCaCl)/(1 mol CaCL# = $19.99 g$ $C a C l$