How many grams of calcium chloride (CaCl2; molecular weight = 111 g mole^−1) would you dissolve in water to make a 0.2 M CaCl2 solution with 200 mL final volume?

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This is a molarity problem, and molarity is defined by the moles of solute in one liter of solution. In this case, the calcium chloride is the solute (thing being dissolved) and the solution is just the calcium chloride dissolved in water. Basically you're talking about how concentrated or diluted the mixture is but in terms of quantitative ratios.

A 1M, or "one molar" solution, would have one mole of solute in one liter of solution. One liter is the same as 1000 milliliters, so you can swap that term out in any situation where you're using milliliters instead of liters worth of solution.

Your target, a 0.2M solution, would have 0.2 moles of solute in one liter of solution. Furthermore, this is a ratio; you're only going to need 200mL of solution instead of a liter, so you'll need less than 0.2 moles of solute in order to match the ratio. The required amount can be determined by setting the two ratios equal to each other, then cross-multiplying, then solving for the unknown.

1. set ratios equal to each other:

`(0.2 mol )/(1000mL )` = `(X mol)/(200mL)`

2. cross-multiply

(0.2mol)(200mL) = (X mol)(1000mL)

3. solve for X

`((0.2 mol)(200mL))/(1000mL)` = X mol

X = 0.04mol solute.

You can confirm by plugging the moles of solute into the milliliters of solution to check that it still equals 0.2M.

Finally, the 0.04 moles of calcium chloride need to be converted into mass. This involves the molar mass (stated above as the molecular weight, where 1 mole weighs 111g.)

`0.04 mol CaCl_2` x `(111g)/(1 mol)`

The mole units cancel out, leaving gram units, and 0.04 x 111 = 4.44g calcium chloride.

Last Updated by eNotes Editorial on February 5, 2020
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