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ChemMedAdv 2016 • Paper 1

Ratio of Molecular Weights from Molarity and Molality Equality

In a given solution, the mole fraction of a solute equals 0.10.1. At a temperature of 298 K298\text{ K}, the molarity of this solution matches its molality. The density of the solution at 298 K298\text{ K} is 2.0 g cm−32.0\text{ g cm}^{-3}. Determine the ratio of the molar masses of the solute and the solvent, (MWsoluteMWsolvent)\left(\frac{MW_{\text{solute}}}{MW_{\text{solvent}}}\right).

Official Numerical Answer9

Topics & Concepts

SolutionsConcentration Terms

Step-by-Step Solution

To find the ratio of the molecular weights of the solute and the solvent, (MWsoluteMWsolvent)\left(\frac{MW_{\text{solute}}}{MW_{\text{solvent}}}\right), we use the definitions of molarity, molality, and density.

Given Data:

  • Mole fraction of solute (xsolutex_{\text{solute}}) = 0.10.1
  • Mole fraction of solvent (xsolventx_{\text{solvent}}) = 1−0.1=0.91 - 0.1 = 0.9
  • Density of solution (dd) = 2.0 g cm−3=2.0 g mL−12.0\text{ g cm}^{-3} = 2.0\text{ g mL}^{-1}
  • Molarity (MM) = Molality (mm)

Step-by-Step Derivation:

1. Expression for Molality (mm):

Consider 1 mole1\text{ mole} of the total solution mixture.

  • Moles of solute (nsoluten_{\text{solute}}) = 0.1 mol0.1\text{ mol}
  • Moles of solvent (nsolventn_{\text{solvent}}) = 0.9 mol0.9\text{ mol}

The mass of the solvent in kilograms is: Mass of solvent (kg)=nsolvent×MWsolvent1000=0.9×MWsolvent1000\text{Mass of solvent (kg)} = \frac{n_{\text{solvent}} \times MW_{\text{solvent}}}{1000} = \frac{0.9 \times MW_{\text{solvent}}}{1000}

Thus, the molality (mm) of the solution is given by: m=nsoluteMass of solvent (in kg)=0.1(0.9×MWsolvent1000)=1000.9×MWsolventm = \frac{n_{\text{solute}}}{\text{Mass of solvent (in kg)}} = \frac{0.1}{\left(\frac{0.9 \times MW_{\text{solvent}}}{1000}\right)} = \frac{100}{0.9 \times MW_{\text{solvent}}}

2. Expression for Molarity (MM):

The total mass of the solution is: Mass of solution (g)=nsolute×MWsolute+nsolvent×MWsolvent=0.1×MWsolute+0.9×MWsolvent\text{Mass of solution (g)} = n_{\text{solute}} \times MW_{\text{solute}} + n_{\text{solvent}} \times MW_{\text{solvent}} = 0.1 \times MW_{\text{solute}} + 0.9 \times MW_{\text{solvent}}

The volume of the solution in liters is: V (in L)=Mass of solutiond×1000=0.1×MWsolute+0.9×MWsolvent2.0×1000V\text{ (in L)} = \frac{\text{Mass of solution}}{d \times 1000} = \frac{0.1 \times MW_{\text{solute}} + 0.9 \times MW_{\text{solvent}}}{2.0 \times 1000}

Thus, the molarity (MM) of the solution is: M=nsoluteV (in L)=0.1(0.1×MWsolute+0.9×MWsolvent2000)=2000.1×MWsolute+0.9×MWsolventM = \frac{n_{\text{solute}}}{V\text{ (in L)}} = \frac{0.1}{\left(\frac{0.1 \times MW_{\text{solute}} + 0.9 \times MW_{\text{solvent}}}{2000}\right)} = \frac{200}{0.1 \times MW_{\text{solute}} + 0.9 \times MW_{\text{solvent}}}

3. Equating Molarity and Molality:

Since the molarity of the solution matches its molality (M=mM = m): 2000.1×MWsolute+0.9×MWsolvent=1000.9×MWsolvent\frac{200}{0.1 \times MW_{\text{solute}} + 0.9 \times MW_{\text{solvent}}} = \frac{100}{0.9 \times MW_{\text{solvent}}}

Dividing both sides by 100100: 20.1×MWsolute+0.9×MWsolvent=10.9×MWsolvent\frac{2}{0.1 \times MW_{\text{solute}} + 0.9 \times MW_{\text{solvent}}} = \frac{1}{0.9 \times MW_{\text{solvent}}}

Cross-multiplying gives: 2×(0.9×MWsolvent)=0.1×MWsolute+0.9×MWsolvent2 \times \left(0.9 \times MW_{\text{solvent}}\right) = 0.1 \times MW_{\text{solute}} + 0.9 \times MW_{\text{solvent}}

1.8×MWsolvent=0.1×MWsolute+0.9×MWsolvent1.8 \times MW_{\text{solvent}} = 0.1 \times MW_{\text{solute}} + 0.9 \times MW_{\text{solvent}}

Rearranging the terms to collect MWsolventMW_{\text{solvent}} on one side: (1.8−0.9)×MWsolvent=0.1×MWsolute(1.8 - 0.9) \times MW_{\text{solvent}} = 0.1 \times MW_{\text{solute}}

0.9×MWsolvent=0.1×MWsolute0.9 \times MW_{\text{solvent}} = 0.1 \times MW_{\text{solute}}

Therefore, the required ratio is: MWsoluteMWsolvent=0.90.1=9\frac{MW_{\text{solute}}}{MW_{\text{solvent}}} = \frac{0.9}{0.1} = 9

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