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Chapter 13: Problem 53

Calculate the vapor pressure at \(25^{\circ} \mathrm{C}\) of a solution containing \(165 \mathrm{g}\) of the nonvolatile solute, glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in \(685 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(23.8 \mathrm{mmHg}\).

Short Answer

Expert verified
The vapor pressure of the solution at \(25^{\circ} C\) is \(23.2 mmHg\).

Step by step solution

01

Calculate the moles of glucose and water

To calculate the mole fraction, it's first required to know the moles of glucose and water. Use the given mass for each and their molar masses for this. For glucose \(C_6H_{12}O_6\), the molar mass is approximately \(180.156 g/mol\). For water \(H_2O\), it's \(18.015 g/mol\). Thus, \n\nMoles of glucose = \(165g / 180.156 g/mol = 0.916 mol\) \n\nMoles of water = \(685g / 18.015 g/mol = 38 mol\)
02

Calculate the mole fraction of water

The mole fraction is calculated by dividing the moles of water by the total moles in the solution. It is given by \n\nMole fraction of water = Moles of water / (Moles of water + Moles of glucose) \n\nSubstituting the values, \n\nMole fraction of water = \(38 mol / (38 mol + 0.916 mol) = 0.976\)
03

Calculate the vapor pressure of the solution

Finally, apply Raoult's law, which states that the vapor pressure of a solution is equal to the mole fraction of the solvent times the vapor pressure of the pure solvent. \n\nThe vapor pressure of the solution \( P_{solution}\) is thus given by \n\n\( P_{solution} \) = Mole fraction of water \(\times \) Vapor pressure of pure water \n\nSubstituting the values, \n\n\( P_{solution} \) = \(0.976 \times 23.8 mmHg = 23.2 mmHg \)

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Most popular questions from this chapter

Which of the following ions has the greater charge density? (a) \(\mathrm{Na}^{+} ;\) (b) \(\mathrm{F}^{-} ;\) (c) \(\mathrm{K}^{+} ;\) (d) \(\mathrm{Cl}^{-}\).

At 1.00 atm, the solubility of \(\mathrm{O}_{2}\) in water is \(2.18 \times 10^{-3} \mathrm{M}\) at \(0^{\circ} \mathrm{C}\) and \(1.26 \times 10^{-3} \mathrm{M}\) at \(25^{\circ} \mathrm{C}\) What volume of \(\mathrm{O}_{2}(\mathrm{g}),\) measured at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm},\) is expelled when \(515 \mathrm{mL}\) of water saturated with \(\mathrm{O}_{2}\) is heated from 0 to \(25^{\circ} \mathrm{C}\) ?

Which of the following do you expect to be most water soluble, and why? \(\mathrm{C}_{10} \mathrm{H}_{8}(\mathrm{s}), \mathrm{NH}_{2} \mathrm{OH}(\mathrm{s})\) \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l}), \mathrm{CaCO}_{3}(\mathrm{s})\).

The aqueous solubility at \(20^{\circ} \mathrm{C}\) of \(\mathrm{Ar}\) at \(1.00 \mathrm{atm}\) is equivalent to \(33.7 \mathrm{mL} \mathrm{Ar}(\mathrm{g}),\) measured at STP, per liter of water. What is the molarity of Ar in water that is saturated with air at 1.00 atm and \(20^{\circ} \mathrm{C}\) ? Air contains \(0.934 \%\) Ar by volume. Assume that the volume of water does not change when it becomes saturated with air.

A water sample is found to have 9.4 ppb of chloroform, \(\mathrm{CHCl}_{3} .\) How many grams of \(\mathrm{CHCl}_{3}\) would be found in a glassful \((250 \mathrm{mL})\) of this water?
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