Question

2.50 g of NH4Cl is added to 12.9 g of water. Calculate the molality of the solution.

Answer 1

Molality(m) of a solution is defined as the number of moles of solute present in 1 kg of solvent.

According to question, NH₄Cl is the solute and water, that is, H₂O is the solvent.

Given Mass of solute, m_solute = 2.50 g

Atomic Mass of N - 14.007 u

Atomic Mass of H - 1.008 u

Atomic Mass of Cl - 35.45 u

So, Molar Mass of solute, M_solute = (14.007)1 + (1.008)4 + (35.45)1 g/mol = 53.49 g/mol

So, Number of moles of solute, n = m_solute/M_solute = (2.50 g)/(53.49 g/mol) = 0.0467 mol

Weight of the solvent, W = 12.9 g

1 kg = 1000 g

So, 1 g = 10^-3 kg

Weight of the solvent, W = (12.9 g)(10^-3 kg/1 g) = 12.9 x 10^-3 kg = 0.0129 kg

Therefore, the Molality of the solution, m = n/W = (0.0467 mol)/(0.0129 kg) = 3.62 mol kg^-1

1 molal = 1 mol kg^-1

Therefore,

m = 3.62 molal

Therefore, the molality of the given solution is 3.62 molal.