Question

A solution is made by mixing 10.2 grams of CaCl2 in 250 grams of water. What is the molality of the

solution?

Answer 1

0.367 m

Step-by-step explanation:

We have to start with the molality equation:

`m=(mol~of~solute)/(Kg~of~solvent)`

If we want to calculate the molality we need the moles of the solute and the Kg of the solvent. In this case the solute is
`CaCl_2` and the solvent is
`H_2O`.

Lets start with the the moles of solute. For this calculation we need the molar mass of
`CaCl_2` (
`111~g~CaCl_2`) and the mass, so:

`10.2~g~CaCl_2(1~mol~CaCl_2)/(111~g~CaCl_2)`

`0.0918~mol~CaCl_2`

Now, we can calculate the Kg of solvent, if we do the conversion from grams to Kg:

`250~g~CaCl_2(1~Kg)/(1000~g)`

`0.25~Kg`

With these values we can calculate the "molality":

`m=(0.0918~mol~CaCl_2)/(0.25~Kg)`

`m=0.3672`

I hope it helps!

Answer 2

The molality of this solution is 0.368 molal

Step-by-step explanation:

Step 1: Data given

Mass of CaCl2 = 10.2 grams

Molar mass of CaCl2 = 110.98 g/mol

Mass of water = 250 grams = 0.250 kg

Step 2: Calculate the number of moles CaCl2

Moles CaCl2 = mass CaCl2 / molar mass CaCl2

Moles CaCl2 = 10.2 grams / 110.98 g/mol

Moles CaCl2 = 0.0919 moles

Step 3: Calculate the molality of the solution

Molality solution = moles CaCl2 / mass water

Molality solution = 0.0919 moles / 0.250 kg

Molality solution = 0.368 mol / kg = 0.368 molal

The molality of this solution is 0.368 molal