When water and methanol, \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l}),\) are mixed, the total volume of the resulting solution is not equal to the sum of the pure liquid volumes. (Refer to Exercise 99 for an explanation.) When \(72.061 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) and \(192.25 \mathrm{g}\) \(\mathrm{CH}_{3} \mathrm{OH}\) are mixed at \(25^{\circ} \mathrm{C},\) the resulting solution has a density of \(0.86070 \mathrm{g} / \mathrm{mL} .\) At \(25^{\circ} \mathrm{C},\) the densities of water and methanol are \(0.99705 \mathrm{g} / \mathrm{mL}\) and \(0.78706\) \(\mathrm{g} / \mathrm{mL},\) respectively. (a) Calculate the volumes of the pure liquid samples and the solution, and show that the pure liquid volumes are not additive. [ Hint: Although the volumes are not additive, the masses are.] (b) Calculate the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in this solution.

Step by step solution

01

Calculate the volume of component liquids

Using the formula \( Volume = \frac{Mass}{Density} \), calculate volume of water (\(V_{H2O}\)) as \( \frac{72.061 g}{0.99705 g/mL} \) and volume of methanol (\(V_{CH3OH}\)) as \( \frac{192.25 g}{0.78706 g/mL} \).

02

Calculate total volume of the solution

The total mass of the solution is the sum of the masses of water and methanol i.e. \( 72.061 g + 192.25 g \). Now, using this total mass and the density of the solution, calculate the total volume of the mixture (\(V_{mixture}\)). Use the formula \( Volume = \frac{Mass}{Density} \), so \(V_{mixture}\) = \( \frac{72.061 g + 192.25 g}{0.86070 g/mL} \).

03

Compare the volumes

Now, compare \(V_{H2O} + V_{CH3OH}\) and \(V_{mixture}\). Show that these two volumes are not equal which means the volumes of pure liquids are not additive in mixture.

04

Calculate moles of CH3OH

To calculate the molarity, first, calculate the number of moles of methanol using its given mass and molar mass. The molar mass of \(CH_3OH\) is \(32.04 g/mol\). So, moles of \(CH_3OH\) = \( \frac{192.25 g}{32.04 g/mol} \).

05

Calculate molarity of CH3OH

The molarity of \(CH_3OH\) is defined as the number of moles of \(CH_3OH\) per liter of solution. First convert the volume of the solution (from step 2) from mL to L by dividing by 1000. Now, calculate the molarity as \( \frac{moles of CH_3OH}{Volume of solution in L} \).

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