At 20 °C,a 2.72 mol L^-1 aqueous solution of ammonium chloride has a density of $1.0687 \mathrm {~g} \mathrm {~mL} ^ { - 1 }$ What is the molality of ammonium chloride in the solution? The formula weight of $\mathrm { NH } _ { 4 }$ Cl is $53.50 \mathrm {~g} \mathrm {~mol} ^ { - 1 }$ A)0.339 mol kg^-1 B)2.72 mol kg^-1 C)0.393 mol kg^-1 D)13.6 mol kg^-1 E)2.95 mol kg^-1

Question

Quizplus · Accepted Answer

To find molality, we need the moles of solute per kilogram of solvent. First, calculate the mass of the solution: 1 L of solution has a mass of 1068.7 g. The mass of NH4Cl in 1 L is 2.72 mol × 53.50 g/mol = 145.52 g. The mass of water is 1068.7 g - 145.52 g = 923.18 g = 0.92318 kg. Molality = 2.72 mol / 0.92318 kg = 2.95 mol/kg.

At 20 °C,a 2 1.0687 g mL−11.0687 \mathrm {~g} \mathrm {~mL} ^ { - 1 }1.0687 g mL−1

At 20 °C,a 2 $1.0687 \mathrm {~g} \mathrm {~mL} ^ { - 1 }$