Converting Between Concentration Units

Density to the rescue! Molarity and Molality Often a laboratory will provide you with the concentration of the solution in molarity, but the colligative property needs a different unit.

The information most often utilized for converting between molarity and molality is density (g/ml). Example Problem (M -> m)

Find the molality of 18 M H2SO4. This solution has a density of 1.84 g/mL.

Step 1: make an assumption that you have 1 L of the solution. Step 2: with that assumption made, you can them multiply the density by 1 L to determine the amount of grams of solution 1 L | 1000 mL | 1.84 g = 1L 1 mL

= 1840 g of solution Example Problem (M -> m)

Step 3: Now you need to determine how many grams of H2SO4 were present. You assumed you have 1 L of solution, so that means you have 18 moles of H2SO4. Convert 18 moles of H2SO4 to grams.

18 mol H2SO4 | 98.08 g H2SO4 = 1 mol H2SO4 = 1765. 62 g H2SO4

Example Problem (M -> m)

Step 4: now determine how many grams of solvent

1840 g of solution - 1765.62 g of H2SO4 = 74.38 g solvent Step 5: determine molality

18 mol H2SO4 = .07439 kg

= 242 m H2SO4 Practice

A solution of sodium nitrate has a concentration of 0.733 M. Its density is 1.0392 g/mL. What is the molality?

Answer:

NaNO3 (sodium nitrate) Step 1: have 1L of solution Step 2: 1L *(1000mL/1L) * (1.0392g/1mL) = 1039.2g

Step 3: 0.733 mol NaNO3 * (84.995g/1mol) = 62.301 g NaNO3 Step 4: 1039.2g – 62.301 g = 976.9g

Step 5: 0.733 mol NaNO3/ 0.9769 kg = 0.750 m NaNO3

Example Problem (m -> M)

An aqueous solution of Na2SO4 has a concentration of 0.370 m. Its density is 1.0436 g/mL. What is the Molarity? Step 1: Assume that you have 1 kg of solvent

Step 2: Determine the amount of Na2SO4 you have

0.370 mol Na2SO4 | 142.04 g Na2SO4 =

1 mol Na2SO4

= 52.55 g Na2SO4 Example Problem (m -> M)

Step 3: calculate the total grams of solution

1000 g + 52.55 g Na2SO4 = 1052.55 g Step 4: convert to L using density 1052.55 g | 1 mL | 1 L = 1.0436 g | 1000 ml = 1.01 L Step 5: solve for molarity

.370 mol Na2SO4/ 1.01 L = = 0.366 M Na2SO4 Problem

Find the molarity of 21.4 m HF. This aqueous solution has a density of 1.101 g/mL. Answer

Step 1: 1 kg of solution Step 2: 21.4 mol HF * 20.01 g/mol = 428.21 g HF Step 3: 1000 g solvent (water) + 428.21 g HF = 1428.21 g solution Step 4: 1428.21 g * 1mL/1.101 g (density) = 1297.19 mL = 1.29719 L Step 5: 21.4 mol HF / 1.29719 L = 16.5 M HF