This document discusses stoichiometry calculations involving molarity and molality. It provides examples of calculating molarity, molality, mass percent, and mole fraction for various solutions. Molarity is moles per liter of solution, while molality is moles per kilogram of solvent. Examples include calculating the molality of NaCl in water, the molarity of HCl in a solution, and the molality, molarity, and mass of various solutes needed to make solutions of specific concentrations.

3.  Stoichiometry Calculations
It includes- Numerical based on
Molarity and Molality

4. In the last part we have seen about limiting reagent, Mass
percent and mole fraction. The following are the formula for
the above:

6. The important part of remembering the difference is:
molarity - M → moles per liter solution
molality - m → moles per kilogram solvent

7. 25.0 g of NaCl is dissolved in 5000 ml of
water. Find the molality (m) of the resulting
solution.
 m = mol solute/ kg solvent
 Since, the density of water is 1g/ml,
 5000ml = 5000 g, which is 5 kg
 0.427 mol NaCl / 5 kg water = 0.0854 m salt water
n = m /GMM

8. What is the molarity of HCl if 28g HCl
is dissolved in 500.0 mL of solution?

10. Calculate the molality of a solution prepared from
29.1 g of toluene, C7H8 dissolved in 832 g of
benzene, C6H6
 Molar Mass of toluene i.e. C7H8 = 92 g/mol
 Moles of toluene = mass/ molar mass
= 29.1 g/ 92.0 g/mol
= 0.316 mol
 m = mol solute/ kg solvent
 = 0.316/0.832 kg
= 0.380 m

12. An aqueous solution of HCl is 30.0% HF, by mass, and has a
density of 1.101 g cm3. what are the molality and
molarity of HF in this solution
 3) Calculate the molarity:
 We got moles of HF = 1.49956mol
 Determine volume of 100g of solution:
Density = mass/volume
=1.101 g/mL = 100.0 g/x
X = 90.8265 mL
 Determine Molarity:
M = moles of solute/ volume in L
= 1.49956/0.0908265 L
= 16.5 M

13. Using Molarity
What mass of oxalic acid (C2H2O4)
required to make 250 mL of a 0.0500 M
solution?
 Molar mass of C2H2O4 = 90.00 g/mol
 Step 1: Change mL to L
250 mL * 1L/ 1000 mL = 0.250 L
 Step 2: Calculate moles: Moles = M * V
Moles = (0.0500 mol/L) X (0.250 L)
= 0.0125 moles
 Step 3: Convert moles to grams
 n = mass/molar mass,
 Mass = n X molar mass
=(0.0125 mol) X (90.00 g/mol)
=1.125 g

16. Solution:

21.  7.45 g of potassium chloride is dissolved in 100 g of water. What will be the
molality of the solution?
 1.11 g of urea was dissolved in 100 g of water. Calculate the molality of
solution. (N=14, H=1. C=12, O=16]