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Chapter 12: Problem 94

A solution of \(1.00 \mathrm{~g}\) of anhydrous aluminum chloride, \(\mathrm{AlCl}_{3}\), in \(50.0 \mathrm{~g}\) of water freezes at \(-1.11^{\circ} \mathrm{C}\). Does the molar mass determined from this freezing point agree with that calculated from the formula? Why?

Short Answer

Expert verified
The molar mass determined from both the freezing point depression and the formula should agree if the conditions are ideal. Any discrepancy might be due to phenomena like ion pairing in solution.

Step by step solution

01

Calculate the molality

The molality (m) can be calculated using the formula \(m = \frac{mass~of~solvent~(kg)}{moles~of~solute}\). Given that the mass of the solvent (water) is 50.0 g or 0.050 kg and the mass of solute (aluminum chloride) is 1.00 g, you first need to convert the gram of AlCl3 to moles using its molar mass (133.34 g/mol). So, \(m = \frac{0.050}{1.00/133.34}\).
02

Calculate the depression in freezing point

The depression in freezing point \(\Delta T_f\) can be calculated using the formula \(\Delta T_f = i * K_f * m\), where 'i' is the van't Hoff factor (for AlCl3 it is 4), and 'K_f' is the freezing point depression constant equal to 1.86 °C/m for water. So, \(\Delta T_f = 4 * 1.86 * m\).
03

Compare the experimental freezing point depression with the calculated one

Doing the step 2 calculations gives the expected freezing point depression of water when 1.00 g of AlCl3 is dissolved. You then compare this result to the given experimental freezing point depression of -1.11 °C. If they match, the molar mass determined from the experimental freezing point agrees with that calculated from the formula. If there's a discrepancy, an explanation needs to be searched, like ion pairing that reduces the effective concentration of particles.

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Most popular questions from this chapter

Use a phase diagram to show the difference in freezing points and boiling points between an aqueous urea solution and pure water.

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