A solution consists of 44.5 g of silver nitrate, AgNO₃, and 106.5 g water.

Calculate the weight percent, the molality, and the mole fraction of AgNO₃in the solution.

Weight percent = 29.5%

Molality = 2.460 mol/kg

Mole fraction = 0.042

Explanation:

The weight percent (%w) of a substance is the mass of it divided by the total mass of the solution, and then multiplied by 100%, so:

%w = [44.5 / (44.5 + 106.5) ] * 100%

%w = 29.5%

The molality (W) of a solution is the number of moles of the solute divided by the mass (in kg) of the solvent. In this case, water is the solvent, and AgNO₃ the solute (water is generally the solvent, and the solvent is always in a larger amount). The number of moles is the mass divided by the molar mass, and the molar mass of AgNO₃ is 169.87 g/mol:

n = 44.5/169.87 = 0.2620 mol

W = 0.2620/0.1065 = 2.460 mol/kg

The mole fraction (x) of a substance is the number of moles of it divided by the total number of moles of the solution. The molar mass of water is 18 g/mol, so the number of moles of water is:

n = 106.5/18 = 5.92 moles

Thus the mole fraction of AgNO₃ is:

x = 0.2620 / (0.2620 + 5.92) = 0.042