Question

An aqueous solution is 36% by mass silver nitrate, AgNO3, and has a density of 1.41 g/mL. What is the molality of silver nitrate in the solution?

Answer 1

To find the molality of the silver nitrate solution, we can use the mass percent and density of the solution. The molality is found to be 2.99 mol/kg.

Step-by-step explanation:

To find the molality of the silver nitrate solution, we need to first determine the moles of silver nitrate in the solution. We can use the mass percent and density to do this.

First, let's assume we have 100 g of the solution.

This means we have 36 g of silver nitrate in the solution (since it is 36% by mass).

The molar mass of AgNO3 is 169.88 g/mol, so we can convert the mass of the silver nitrate to moles: 36 g / 169.88 g/mol = 0.2124 mol.

The density of the solution is given as 1.41 g/mL, so 100 g of the solution would have a volume of 100 g / 1.41 g/mL = 70.92 mL.

Now, we can calculate the molality using the moles of AgNO3 and the mass of the water:

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

Molality = 0.2124 mol / 0.07092 kg

= 2.99 mol/kg