Question

An aqueous solution of glycerin, 7% by weight, is prepared. The solution is found to have a density of 1.0149 g/cm³ at 20C. The molecular weight of glycerin is 92.0473 g/mol and its density is 1.2609 g/cm³ (or g/mL) at 20C. What are the molarity, molality, and percent by volume of this solution?

Answer 1

To calculate the molarity, molality, and percent by volume of a 7% by weight glycerin solution with a specified density, we find the mass of glycerin, convert it to moles, and use the density to determine the solution's volume. Molarity is found by dividing moles by the volume of the solution, while molality is moles per kilogram of solvent, and percent by volume is derived from the glycerin volume over the total solution volume.

Step-by-step explanation:

Calculation of Molarity, Molality, and Percent by Volume

Firstly, we can compute the mass of glycerin in 1000 g of the solution since the solution is 7% by weight. Mass of glycerin = 7% of 1000 g = 70 g. Next, we can find the volume of glycerin using its density: Volume of glycerin = mass / density = 70 g / 1.2609 g/mL = 55.52 mL.

To calculate the molarity of the glycerin solution, divide the moles of glycerin by the volume of the solution in liters. Number of moles = mass / molecular weight = 70 g / 92.0473 g/mol = 0.7606 mol. Assuming the solution's volume is approximately that of water, its volume is 1 liter (since the density of the solution is close to 1 g/mL), so the molarity is 0.7606 M.

For molality, we calculate moles of solute per kilogram of solvent. We previously found the moles of glycerin to be 0.7606 mol. The mass of water in the solution is 930 g (1000 g of solution - 70 g of glycerin), which is 0.93 kg. Thus, molality is 0.7606 mol / 0.93 kg solvent = 0.8178 m.

The percent by volume is given by the volume of glycerin divided by the total volume of the solution. Assuming 1 mL = 1 g for the solution's density, the total volume is 1000 mL. Thus, percent by volume is 55.52 mL / 1000 mL * 100 = 5.552%.