Question

An aqueous solution of non-electrolyte A, with a molecular
mass of 60, contains 6 grams of non-electrolyte A in 500 mL
and has a density equal to 1.05 g per mL. The molality of the
solution is

Answer 1

Answer:0.19m

Explanation:

Answer 2

Molality (m) of the solution is  = 0.192 mol/kg. Molality (also molal concentration) is represented by symbol m, it measures the number of moles of solute (here a non-electrolyte solute A) present  in 1 kg of solvent. Here, moles of solute can be calculated by dividing given mass of non-electrolyte A with molecular weight of A.

Molecular mass of non-electrolyte A = 60g

Mass of non-electrolyte A = 6g

Volume of solution = 500mL

Density of solution = 1.05g

Calculating molality of solution

To calculate molality of solution using formula of molality:

Molality = moles of solute / kilograms of solvent (mol/kg). Here we need to calculate mass of the solvent in kg that is not directly given in the question. Hence, it is required to calculate mass of the solution first and then subtract the mass of solute from it. Also, to calculate the mass of solution, density and volume are given. So, the calculations are done as follows;

Firstly,

To calculate mass of solution using formula of density; D (ρ) = mass/volume. Density (ρ) is the mass of substance divide by it's volume. Therefore, Mass of solution = Density  × volume

Mass of solution = 1.05 × 500

Mass of solution = 525 g

Now, Mass of solvent is calculated as Mass of solute (non-electrolyte A) subtracted from mass of solution. Therefore,

Mass of solvent = 525 - 6 = 519 g (0.519 kg)

Moles of solute are calculated as given weight of solute divided by molecular weight of solute. Hence, moles of solute = 6/60

Moles of solute = 0.1

Molality of solution is calculated as moles of solute divided by mass of solvent or kilograms of solvent (mol/kg)

Molality of solution = 0.1/ 0.519

Molality (m) = 0.192 mol/kg

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