Molality (m) – the number of moles of solute dissolved in 1 Kg of solvent.
Molality (m) =
moles of solute
Kg of Solvent
Examples:
1)
Find the Molality of a solution made with 15.3 g of NH4Cl in 0.835 Kg of water.
15.3 g x
m=
2)
1 mol
= 0.286 mol
53.4912 g
mol solute
0.286 mol
=
= 𝟎. 𝟑𝟑𝟓 𝐦
Kg solvent
0.835 Kg
How many moles and grams of NaOH are needed to make a 4.35 m solution of NaOH
using 525 g of water?
Since m =
mol
then mol = m x Kg = 4.35 m x 0.525 Kg = 𝟐. 𝟐𝟖 𝐦𝐨𝐥 𝐍𝐚𝐎𝐇
Kg
2.28 mol x
39.9971 g
= 𝟗𝟏. 𝟐 𝐠 𝐍𝐚𝐎𝐇
1 mol
% Concentration by mass:
% 𝐜𝐨𝐧𝐜. 𝐛𝐲 𝐦𝐚𝐬𝐬 =
𝐦𝐚𝐬𝐬 𝐬𝐨𝐥𝐮𝐭𝐞
𝐱 𝟏𝟎𝟎
𝐦𝐚𝐬𝐬 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
Examples:
1)
75.4 g of Ethylene glycol are dissolved in 250.0 g of water. Calculate the % conc.
Mass solute = 75.4 g
Mass solution = mass solute + mass solvent = 75.4 g + 250.0 g = 325.4 g solution
% conc. by mass =
mass solute
75.4 g
x 100 =
x 100 = 𝟐𝟑. 𝟐 %
mass solution
325.4 g
2)
500.0 g of a 62.3 % solution of Sodium sulfate is made, how much Sodium sulfate
was used?
(% conc. )(mass solution)
(62.3 %)(500.0 g)
=
= 𝟑𝟏𝟐 𝐠 𝐒𝐨𝐝𝐢𝐮𝐦 𝐬𝐮𝐥𝐟𝐚𝐭𝐞
100
100
For a 458 mL solution with 36.5% HCl by mass and a density of 1.19 g/ml. Find:
mass solute =
3)
A) mass of the solution
458 mL x
1.19 g
= 𝟓𝟒𝟓 𝐠 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
1 mL
B) mass of HCl
mass solute =
(% conc. )(mass solution)
(36.5 %)(545 g)
=
= 𝟏𝟗𝟗 𝐠 𝐇𝐂𝐥
100
100
C) moles of HCl
199 g HCl x
1 mol HCl
= 𝟓. 𝟒𝟒 𝐦𝐨𝐥 𝐇𝐂𝐥
36.56094 g HCl
D) Molarity of the solution
Molarity =
mol solute
5.44 mol
=
= 𝟏𝟏. 𝟗 𝐌
Liters solution
0.458 L
E) Molality of the solution
Kg Solvent = Kg Solution – Kg Solute = .545 kg - .199 kg = 0.346 Kg
Molality =
mol solute
5.44 mol
=
= 𝟏𝟓. 𝟕 𝐦
Kg solvent
0. .346 Kg
Solution Stoichiometry:
1) How many mL of 0.100 M NaOH are required to neutralize 50.0 mL of 0.250 M H2SO4?
2 NaOH(aq) + H2SO4(aq)

Na2SO4(aq)
+
2 H2O(l)
Plan: Determine the number of moles of H2SO4 present. Use the molar ratio to find the
number of moles of NaOH required. Determine the volume needed.
mol = Molarity x Volume = (0.250 M)(0.050 L) = 00125 mol H2SO4
0.0125 mol H2 SO4 =
V=
2 mol NaOH
= 0.0250 mol NaOH
1 mol H2 SO4
mol
0.0250 mol NaOH
1000 mL
=
= 0.250 L NaOH x
= 𝟐𝟓𝟎. 𝐦𝐋 𝐍𝐚𝐎𝐇
Molarity
0.100 M NaOH
1L
2) If 252 mL of 1.05 M Phosphoric acid is added to a Magnesium hydroxide solution, how
much Magnesium phosphate will be produced?
2 H3PO4(aq)
+
3 Mg(OH)2(aq) 
Mg3(PO4)2(aq)
+
6 H2O(l)
Mg3(PO4)2 = 262.8578 g/mol
1.05
𝑚𝑜𝑙𝐻3 𝑃𝑂4
1 𝑚𝑜𝑙 𝑀𝑔3 (𝑃𝑂4 )2 262.8578𝑔 𝑀𝑔3 (𝑃𝑂4 )2
𝑥 0.252 𝐿 𝑥
𝑥
= 34.8 𝑔 𝑀𝑔3 (𝑃𝑂4 )2
𝐿 𝐻3 𝑃𝑂4
2 𝑚𝑜𝑙 𝐻3 𝑃𝑂4
1𝑚𝑜𝑙 𝑀𝑔3 (𝑃𝑂4 )2
3) If 252 mL of 1.29 M Magnesium hydroxide solution is added to Phosphoric acid, how
much Magnesium phosphate will be produced?
1.29
𝑚𝑜𝑙 𝑀𝑔(𝑂𝐻)2
1 𝑚𝑜𝑙 𝑀𝑔3 (𝑃𝑂4 )2 262.8578𝑔 𝑀𝑔3 (𝑃𝑂4 )2
𝑥 0.252 𝐿 𝑥
𝑥
= 28.5 𝑔 𝑀𝑔3 (𝑃𝑂4 )2
𝐿 𝑀𝑔(𝑂𝐻)2
3 𝑚𝑜𝑙 𝑀𝑔(𝑂𝐻)2
1𝑚𝑜𝑙 𝑀𝑔3 (𝑃𝑂4 )2
4) If 252 mL of 1.05 M Phosphoric acid is added to 252 mL of 1.29 M Magnesium hydroxide
solution, how much Magnesium phosphate will be produced?
Based on the calculations in problems 2 and 3 the Phosphoric acid can produce 34.8 g, but the
Magnesium hydroxide can only produce 28.5 g. Therefore, the reaction will stop once 28.5 g
are produced since the magnesium hydroxide would be used up at that point, even though
there is still Phosphoric acid left.
The magnesium hydroxide is the limiting reactant while the Phosphoric acid is the excess
reactant.