# How many grams of CaCl2 would you have to add to 235 g of water to make a solution with a concentration of 2.2 M?

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We need to make an aqueous solution of CaCl2 (calcium chloride) with a concentration of 2.2 M. The "M" means molarity, or a measurement of concentration expressed in moles per liter. So we need to find out how many moles of CaCl2 are required to dissolve in the given amount of water (235 g) to make a concentration of 2.2 M. Since the density of water at room temperature and atmospheric pressure is 1 gram per mL, then 235 g of water is also equal to 235 mL of water, or 0.235 liters of water. So that is our denominator. We need to figure out how many moles of CaCl2 are required and then convert that to grams.

2.2 M CaCl2 = x moles CaCl2 / 0.235 liters

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x = (2.2 moles/liter) * 0.235 liters = 0.517 moles CaCl2

Now we multiply the number of moles by the molecular weight of CaCl2 to get the amount in grams.

0.517 moles CaCl2 * (110.98 g/mole) = 57.38 g CaCl2

**So 57.38 g of CaCl2 would need to be dissolved in 235 g of water to make a 2.2 M solution.**

The equation for molarity is:

`M = "mol"/"volume"`

Since water's density is 1 g = 1 mL, there are 235 mL of water.

`235 "mL" = (1 "L")/(1000 "mL") = 0.235 "L"`

Plug in your known values to the first equation:

`2.2 "M" = "mol"/(0.235 "L")`

Solve for number of moles.

`"mol" = 0.517`

Because you need the answer in grams, you must use stoichiometry.

`0.517 "mol" = (110.98 "g")/(1 "mol") = 57.38 "grams"`

Don't forget to put the answer in two sig figs!