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
|1L
=
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