Question

2. A solution of NH₃ in water has a concentration of 4.48%. Calculate the molality of the solution.

3. What is the molality of NaCl in a solution that is 4.000 M NaCl, with a density of 1.55 g mL^-1?

Answer 1

1)   Suppose we have 100 g of solution.
(4.48 g NH3)(mol NH3 / 17.0307 g NH3) = 0.263 mol NH3

(100 - 4.48 g H2O)(mol H2O / 18.0148 g H2O) = 5.302 mol H2O

X = (0.263 mol) / (0.263 + 5.302 mol)
X = 0.0473

m = (0.263 mol) / [(100 - 4.48 g)(kg / 1000 g)]
m= 2.75mol/kg

2) molality= mol of solute/kg of solvent

4.00 M means 4.00 moles NaCl in 1 L of solution

Mass of 1 L = 1550 g

Mass of 4.00 moles of NaCl = 4.00 x 58.4428 g/mol= 233.77 g

Mass water = 1550 - 233.77 = 1316.23 g => 1.316 Kg

m = 4.00 / 1.316 = 3.039