Question

Calculate the molality of an aqueous solution that is 13.5% by mass calcium chloride. You might need to know that the density is 1.24 g/mL A. 1.216 B. 1.765 OC. 1.508 D. 1.406 OE. 2.067

Answer 1

The molality of the aqueous solution that is 13.5% by mass calcium chloride is approximately 0.979 mol/kg.

Step-by-step explanation:

To calculate the molality of the aqueous solution, we need to first calculate the mass of calcium chloride present in 100 g of the solution. Since the solution is 13.5% by mass calcium chloride, we can calculate:

Mass of calcium chloride = (13.5/100) * 100 g = 13.5 g

Next, we need to calculate the mass of water in the solution. Since the density of the solution is 1.24 g/mL, the mass of 100 mL of solution is 1.24 * 100 = 124 g.

Now, we can calculate the molality using the formula:

Molality = (moles of solute) / (mass of solvent in kg)

moles of calcium chloride = (13.5 g) / (110.98 g/mol) = 0.1213 mol

mass of solvent in kg = (124 g) / 1000 = 0.124 kg

Molality = (0.1213 mol) / (0.124 kg) = 0.979 mol/kg

Therefore, the molality of the solution is approximately 0.979 mol/kg.

Answer 2

D. 1.406 m

Step-by-step explanation:

Hello,

At first we need a basis, that could be:
`m_(solution)=100g`, secondly, based on the initial basis, to get the 13.5% by mass, it implies that the mass of the solute (calcium chloride) is 13.5g:

`m_solute=m_(solution)*m/m=100g*0.135=13.5g_(solute)`

Thus, the mass of the solvent in kg is:

`m_(solvent)=m_(solution)-m_(solute)=100g-13.5g=86.5g*(1kg)/(1000g) \\m_(solvent)=0.0865kg`

Now, the moles of the solute:

`molCaCl_2=13.5g*(1molClCl_2)/(110.68g CaCl_2) =0.122mol CaCl_2`

Finally, the molality results:

`m=(mol_(solute))/(m_(solvent))=(0.122mol)/(0.0865kg)\\\\m=1.41m`

Which pretty close to the D answer.

Best regards.