Molality and mole fraction: Class 12 practice
Molality m = moles of solute ÷ kg of solvent (temperature independent). Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Mole concept
The idea behind molality and mole fraction
Molality m = moles of solute ÷ kg of solvent (temperature independent). Mole fraction x = moles of a component ÷ total moles; all mole fractions add to 1.
Step-by-step method
- Find moles of solute (mass ÷ molar mass).
- Molality = moles ÷ kg solvent.
- Mole fraction = n_component ÷ n_total.
Worked example 1 (Basic)
162 g of glucose is dissolved in 5000 g of water. Find the molality.
- n = 162/180 = 0.9 mol
- m = 0.9/5 = 0.18 mol/kg
- Answer: 0.18 m
Worked example 2 (Standard)
32 g of NaOH is dissolved in 4000 g of water. Find the molality.
- n = 32/40 = 0.8 mol
- m = 0.8/4 = 0.2 mol/kg
- Answer: 0.2 m
Worked example 3 (Challenge)
Find the mole fraction of NaOH when 36 g is dissolved in 500 g of water.
- n(solute) = 0.9, n(water) = 27.778
- x = 0.9/(0.9 + 27.778) = 0.0314
- Answer: 0.0314
Common mistakes and misconceptions
- Dividing by the solution's mass
- Using grams of solvent
Practice questions
- (Basic) 18 g of urea is dissolved in 500 g of water. Find the molality.
- (Basic) 17.55 g of NaCl is dissolved in 1500 g of water. Find the molality.
- (Standard) 12 g of urea is dissolved in 4500 g of water. Find the molality.
- (Standard) 162 g of glucose is dissolved in 2000 g of water. Find the molality.
- (Challenge) Find the mole fraction of urea when 18 g is dissolved in 3500 g of water.
- (Challenge) Find the mole fraction of NaOH when 20 g is dissolved in 3000 g of water.
Answer key with full working
- Q1: n = 18/60 = 0.3 mol → m = 0.3/0.5 = 0.6 mol/kg → Answer: 0.6 m
- Q2: n = 17.55/58.5 = 0.3 mol → m = 0.3/1.5 = 0.2 mol/kg → Answer: 0.2 m
- Q3: n = 12/60 = 0.2 mol → m = 0.2/4.5 = 0.044 mol/kg → Answer: 0.044 m
- Q4: n = 162/180 = 0.9 mol → m = 0.9/2 = 0.45 mol/kg → Answer: 0.45 m
- Q5: n(solute) = 0.3, n(water) = 194.444 → x = 0.3/(0.3 + 194.444) = 0.0015 → Answer: 0.0015
- Q6: n(solute) = 0.5, n(water) = 166.667 → x = 0.5/(0.5 + 166.667) = 0.0030 → Answer: 0.0030
Syllabus note
This skill is listed under "Solutions" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is molality and mole fraction in?
"Solutions", Class 12 Chemistry (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.