Q.41 1.45 g of sucrose (C12H22O11) is dissolved in 30.0 ml of water. Molality (rounded off to 3 decimals)
of the resulting solution is _____ m.
Sucrose solution molality calculation yields 0.141 m when 1.45 g is dissolved in 30.0 mL water. This precise value, rounded to three decimals, helps chemistry students master concentration units.
What is Molality?
Molality (m) measures solute moles per kilogram of solvent, unlike molarity which uses solution volume. The formula is m=moles of solutekg of solvent. This unit stays constant with temperature, making it useful in colligative properties studies.
Step-by-Step Calculation
Molar mass of sucrose (C₁₂H₂₂O₁₁) is 342 g/mol (12×12 + 22×1 + 11×16). Moles of sucrose = 1.45 g / 342 g/mol = 0.004237 mol. Solvent mass = 30.0 mL water × 1 g/mL = 30 g = 0.030 kg (density 1 g/mL at room temp). Thus, molality = 0.004237 / 0.030 = 0.141 m (rounded to three decimals).
Common Errors Explained
Using total solution mass instead of solvent gives wrong ~0.132 m; molality requires solvent only. Forgetting kg conversion yields 4.237 m, off by factor of 1000. Molarity confusion (volume-based) might suggest ~0.14 M, but ignores density changes. Precise rounding: 0.004237 mol / 0.030 kg = 0.14123, so 0.141 m.
