Question

45.0 g of CaCl2 are dissolved in enough water so that the molarity of the solution is 1.15 M. What is the volume in mL of the solution?

Answer 1

`\boxed {\boxed {\sf 353 \ mL}}`

Step-by-step explanation:

We are asked to find the volume in milliliters of a solution, given the mass of solute and molarity of the solution.

Molarity is a measure of concentration in moles per liter.

`molarity = \frac {moles \ of \ solute}{liters \ of \ solution}`

1. Moles of Solute

We are given the mass of the solute. We must convert the mass to moles using the molar mass (the mass of 1 mole of a substance). These values are found on the Periodic Table. They are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up the molar masses of the individual elements in calcium chloride.

• Ca: 40.08 g/mol
• Cl: 35.45 g/mol

The chemical formula, CaCl₂, has a subscript of 2. There are 2 moles of chlorine in 1 mole of calcium chloride. We must multiply chlorine's molar mass by 2 before adding calcium's molar mass.

• Cl₂ = 35.45 *2 = 70.9 g/mol
• CaCl₂= 40.08 + 70.9 = 110.98 g/mol

Set up a conversion factor using the molar mass.

`\frac {110.98 \ g \ CaCl_2}{1 \ mol \ CaCl_2}`

Multiply by 45.0 grams of calcium chloride.

`45.0 \ g \ CaCl_2 *\frac {110.98 \ g \ CaCl_2}{1 \ mol \ CaCl_2}`

Flip the conversion factor so the units of grams of calcium chloride cancel.

`45.0 \ g \ CaCl_2 *\frac{1 \ mol \ CaCl_2} {110.98 \ g \ CaCl_2}`

`45.0 \ *\frac{1 \ mol \ CaCl_2} {110.98}`

`\frac {45.0}{110.98} \ mol \ CaCl_2= 0.405478465 \ mol \ CaCl_2`

2. Liters of Solution

Now we can find the liters of solution.

`molarity = \frac {moles \ of \ solute}{liters \ of \ solution}`

• molarity = 1.15 mol CaCl₂/L
• moles of solute = 0.405478465 mol CaCl₂
• liters of solution =x

Substitute the values into the formula.

`1.15 \ mol \ CaCl_2/L = (0.405478465 \ mol \ CaCl_2)/(x)`

Cross multiply.

`\frac {1.15 \ mol \ CaCl_2 /L}{1}= (0.405478465 \ mol \ CaCl_2)/(x)`

`{1.15 \ mol \ CaCl_2 /L}*x= {0.405478465 \ mol \ CaCl_2} * 1`

Divide both sides of the equation by 1.15 moles of calcium chloride to isolate the variable x.

`\frac {1.15 \ mol \ CaCl_2/L *x}{1.15 \ mol \ CaCl_2 /L}= (0.405478465 \ mol \ CaCl_2 * 1 )/(1.15 \ mol \ CaCl_2/L )`

`x=(0.405478465 \ mol \ CaCl_2 * 1 )/(1.15 \ mol \ CaCl_2 /L)`

The units of moles of calcium chloride cancel.

`x=(0.405478465 )/(1.15 \ L )`

`x=0.35258997 \ L`

3. Convert to milliliters

There are 1000 milliliters in 1 liter.

`\frac {1000 \ mL}{1 \ L}`

`0.35258997 \ L * (1000 \ mL)/( 1 \ L)`

`0.35258997 * 1000 \ mL=352.58997 \ mL`

The original measurements of mass and molarity have 3 significant figures, so our answer must have the same. For the number we found, that is the ones place. The 5 in the tenth place tells us to round the 2 up to a 3.

`353 \ mL`

The volume of the solution is approximately 353 milliliters.