Question

In the laboratory you are asked to make a 0.294 m chromium(II) chloride solution using 375 grams of water. How many grams of chromium(II) chloride should you add?

Answer 1

13.5 g of chromium (II) chloride (CrCl2).

Step-by-step explanation:

What is given?

Molality (m) = 0.294 m.

Mass of solvent = 375 g = 0.375 kg.

Molar mass of chromium (II) chloride (CrCl2) = 122.8 g/mol.

Step-by-step solution:

Let's remember the concept of molality: The molality (m) of a solution is the moles of solute divided by the kilograms of solvent. Its formula is the following:

`molality\text{ \lparen m\rparen= }\frac{mole\text{s of solute}}{kilogr\text{ams of solvent}}=(mol)/(kg).`

As we have the molality, and the mass of solvent in kg, we just have to solve for 'moles of solute', which in this case, the solute is CrCl2, and replace the given data, like this:

`\begin{gathered} mole\text{s of solute=molality}\cdot kilograms\text{ of solvent,} \\ \text{moles of solute=0.294 m }\cdot\text{ 0.375 kg,} \\ moles\text{ of solute=0.110 moles.} \end{gathered}`

We have 0.110 moles of CrCl2, but the problem is asking for its mass, so let's convert 0.110 moles to grams using the molar mass of CrCl2:

`0.110\text{ moles CrCl}_2\cdot\frac{122.8\text{ g CrCl}_2}{1\text{ mol CrCl}_2}=13.5\text{ g CrCl}_2.`

The answer is that we need to add 13.5 g of chromium (II) chloride (CrCl2).