Question

What is the molality of a solution prepared by dissolving 18.5 g of magnesium nitrate (MW=148.32 g/mol) in 200.0 g of distilled water?

A.0.564 m
B.0.700 m
C.0.624 m
D.5.76 m
E.2.40 m

Answer 1

The molality of the solution can be calculated by dividing the moles of solute by the mass of the solvent in kilograms. In this case, the molality is approximately 0.6235 m.

Step-by-step explanation:

The molality of a solution is defined as the number of moles of solute dissolved in 1 kilogram of solvent. To calculate the molality, we need to first convert the mass of the solute to moles.

Given: Mass of magnesium nitrate (Mg(NO3)2) = 18.5 g, Molar mass of magnesium nitrate = 148.32 g/mol, Mass of water = 200.0 g.

To determine the molality, we divide the moles of magnesium nitrate by the mass of water in kilograms:

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

First, let's calculate the moles of magnesium nitrate:

moles of Mg(NO3)2 = (mass of Mg(NO3)2) / (molar mass of Mg(NO3)2)

moles of Mg(NO3)2 = 18.5 g / 148.32 g/mol = 0.1247 mol

Next, we convert the mass of water to kilograms:

mass of water = 200.0 g = 0.2 kg

Finally, we calculate the molality:

Molality = 0.1247 mol / 0.2 kg = 0.6235 m

Therefore, the molality of the solution is approximately 0.6235 m.