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Chapter 13: Problem 51

Calculate the number of moles of solute present in each of the following aqueous solutions: (a) \(600 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{SrBr}_{2}\), (b) \(86.4 \mathrm{~g}\) of \(0.180 \mathrm{~m} \mathrm{KCl}\), (c) \(124.0 \mathrm{~g}\) of a solution that is \(6.45 \%\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass.

Short Answer

Expert verified
In summary, the number of moles of solute present in each solution is: (a) \(0.150 \mathrm{~mol}\) of SrBr2, (b) \(0.209 \mathrm{~mol}\) of KCl, and (c) \(0.0444 \mathrm{~mol}\) of glucose.

Step by step solution

01

Convert mL to L

To make calculations, we first need to convert the volume from milliliters (mL) to liters (L) using the conversion factor 1 L = 1000 mL: \(600 \mathrm{~mL} * \cfrac{1 \mathrm{~L}}{1000 \mathrm{~mL}} = 0.600 \mathrm{~L}\)
02

Calculate moles of solute

We know that molality (M) is equal to moles of solute per liter of solution. To calculate the number of moles, we will use the formula: Moles of solute = M x Volume of solution in L Moles of SrBr2 = \(0.250 \mathrm{~M} * 0.600 \mathrm{~L}\) Moles of SrBr2 = \(0.150 \mathrm{~mol}\) #(b) 86.4 g of 0.180 m KCl#
03

Calculate moles of solute from mass and molality

Given the molality (m) and mass of the solution, we can find the number of moles of solute (KCl) using the formula: Moles of solute = molality x (mass of solution / molecular weight of solute) Molecular weight of KCl = 39.1 (K) + 35.5 (Cl) = 74.6 g/mol Moles of KCl = \(0.180 \mathrm{~m} * \cfrac{86.4 \mathrm{~g}}{74.6 \mathrm{~g/mol}}\) Moles of KCl = \(0.209 \mathrm{~mol}\) #(c) 124.0 g of a solution that is 6.45% glucose (C6H12O6) by mass#
04

Calculate mass of solute

To find the mass of glucose in the solution, we will use the percentage given: Mass of glucose = (percentage of glucose / 100) x mass of solution Mass of glucose = \(\cfrac{6.45}{100} * 124.0 \mathrm{~g}\) Mass of glucose = \(8.00 \mathrm{~g}\)
05

Calculate moles of solute from mass

To find the number of moles of glucose, we will use the molecular weight of glucose and the mass of glucose in the solution: Molecular weight of glucose = 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol Moles of glucose = \(\cfrac{8.00 \mathrm{~g}}{180.18 \mathrm{~g/mol}}\) Moles of glucose = \(0.0444 \mathrm{~mol}\)

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Most popular questions from this chapter

A sulfuric acid solution containing \(571.6 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per liter of solution has a density of \(1.329 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate (a) the mass percentage, (b) the mole fraction, (c) the molality, (d) the molarity of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in this solution.

A dilute aqueous solution of an organic compound soluble in water is formed by dissolving \(2.35 \mathrm{~g}\) of the compound in water to form \(0.250 \mathrm{~L}\) of solution. The resulting solution has an osmotic pressure of 0.605 atm at \(25^{\circ} \mathrm{C}\). Assuming that the organic compound is a nonelectrolyte, what is its molar mass?

A solution contains \(0.115 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) and an unknown number of moles of sodium chloride. The vapor pressure of the solution at \(30^{\circ} \mathrm{C}\) is 25.7 torr. The vapor pressure of pure water at this temperature is 31.8 torr. Calculate the number of moles of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)

Caffeine \(\left(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\right)\) is a stimulant found in coffee and tea. If a solution of caffeine in chloroform \(\left(\mathrm{CHCl}_{3}\right)\) as a solvent has a concentration of \(0.0500 \mathrm{~m}\), calculate (a) the percent caffeine by mass, (b) the mole fraction of caffeine.

Two beakers are placed in a sealed box at \(25^{\circ} \mathrm{C}\). One beaker contains \(30.0 \mathrm{~mL}\) of a \(0.050 \mathrm{M}\) aqueous solution of a nonvolatile nonelectrolyte. The other beaker contains \(30.0 \mathrm{~mL}\) of a \(0.035 \mathrm{M}\) aqueous solution of \(\mathrm{NaCl}\). The water vapor from the two solutions reaches equilibrium. (a) In which beaker does the solution level rise, and in which one does it fall? (b) What are the volumes in the two beakers when equilibrium is attained, assuming ideal behavior?
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