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What is the value of $x$ if $x$ molal solution of a compound in benzene has a mole fraction of solute equal to 0.2?

I tried calculating the moles of $x$ through the value given but there are two variables, i.e mole of $x$ and mole of benzene so I am stuck.

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  • $\begingroup$ Molal is moles solute per kilogram solvent. The mol fraction is moles solute per mole solvent. In your case, that means you have .2 moles of solute per 1 mole of benzene. What could you do to get that amount of benzene on kilograms instead of moles? $\endgroup$ – Tyberius Jul 26 at 4:58
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Solution 1

Let's assume there is 1 mole total of solvent and solute. With a mole fraction of 0.2 solute, there is 0.2 mole of solute, the rest (0.8 mole) is benzene. We can calculate the mass of 0.8 mole benzene from the molar mass, and express it in kilograms:

$$ m_\mathrm{benzene} = n_\mathrm{benzene} \cdot M_\mathrm{benzene}$$ $$= \pu{0.8 mol} \cdot \pu{78.11 g/mol} = \pu{62.49 g} = \pu{0.06249 kg}$$

The molality $b_\mathrm{solute}$ is defined as the amount of solute divided by the mass of solvent:

$$ b_\mathrm{solute} = \frac{n_\mathrm{solute}}{m_\mathrm{benzene}} = \frac{\pu{0.2 mol}}{\pu{0.06249 kg}} = \pu{3.201 mol/kg}$$

Solution 2

Let's assume there is 1 kg of benzene. That corresponds to 12.80 moles of benzene. If the mole fraction of solute is 0.2 and that of the solvent benzene is 0.8, they are present in a molar ratio of 1 to 4 (for every four molecules of benzene, there is one molecule of solute). So the amount of solute in that sample is 3.201 mole. We are ready to calculate the molality:

$$ b_\mathrm{solute} = \frac{n_\mathrm{solute}}{m_\mathrm{benzene}} = \frac{\pu{3.201 mol}}{\pu{1 kg}} = \pu{3.201 mol/kg}$$

Problems lacking extensive quantities

It is helpful to understand the difference between extensive and intensive quantities. Intensive quantities do not depend on the size of your sample (examples include density, concentration, molar mass, partial pressure, coefficients in the formula of a compound). On the other hand, extensive quantities do depend on how much you have (mass, volume, amount of substance). Often, you define intensive quantities by dividing one extensive quantity by another (example: density is mass divided by volume).

In your question, you want to determine an intensive quantity (here: molality) from intensive quantities only (here: mole fraction). There is a larger set of exercises of that type. If it helps, you can arbitrarily come up with an extensive quantity to solve the problem. You can set the mass of benzene to 1 kg or 2 kg. Then, you can solve the problem. The answer will not depend on the mass of benzene, it cancels out. Other options are setting the amount of solute or the amount of benzene.

There are common exercises asking you for intensive quantities given other intensive quantities:

  1. What is the formula of an oxide given the mass percentages of elements?
  2. What is the density of xenon at STP given its molar mass?
  3. What is the concentration of chloride ions in a solution containing 0.4 mol/L $\ce{MgCl2}$?
  4. What is the partial pressure of dioxygen at STP if an air sample contains 20% of dioxygen?

For all of these, you may assume an arbitrary amount, volume, or mass (but just one) if it helps to solve the problem. Again, the actual value does not matter because it will cancel out.

For more on intensive and extensive, see e.g. https://chemistry.stackexchange.com/a/6538

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  • $\begingroup$ thanks i get it now $\endgroup$ – studious Aug 5 at 15:00

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