Question

47) What is the molality of a solution prepared by dissolving 317 g of CaCl2 into 2.50 kg of water?

Answer 1

`\( 1.1428 \text{ mol/kg} \)`

Step-by-step explanation:

To calculate the molality of a solution, we use the formula:

`\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \]`

Given:

Mass of
`\( \text{CaCl}_2 \)` (solute) = 317 g

Mass of water (solvent) = 2.50 kg

First, we need to find the moles of
`\( \text{CaCl}_2 \)`.

Molar mass of
`\( \text{CaCl}_2 \)`:

= Atomic weight of Ca + 2 × Atomic weight of Cl

= 40.08 g/mol (for Ca) + 2(35.45 g/mol) (for Cl)

= 40.08 g/mol + 70.90 g/mol

= 110.98 g/mol

Now, calculate the moles of
`\( \text{CaCl}_2 \)`:

`\[ \text{moles of } \text{CaCl}_2 = \frac{\text{given mass}}{\text{molar mass}} \]`

`\[ \text{moles of } \text{CaCl}_2 = \frac{317 \text{ g}}{110.98 \text{ g/mol}} \]`

`\[ \text{moles of } \text{CaCl}_2 = 2.857 \text{ moles} \]`

Now, use the molality formula:

`\[ m = \frac{2.857 \text{ moles}}{2.50 \text{ kg}} \]`

`\[ m = 1.1428 \text{ mol/kg} \]`

Therefore, the molality of the solution is
`\( 1.1428 \text{ mol/kg} \)` or approximately
`\( 1.14 \text{ mol/kg} \)`.