/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Calculate the number of moles of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 in each aqueous solution are: (a) \(0.150 \,\text{moles of}\, \mathrm{SrBr}_{2}\) (b) \(0.0155 \,\text{moles of}\, \mathrm{KCl}\) (c) \(0.0444 \,\text{moles of}\, \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Step by step solution

01

Solve for (a) using volume and molarity

To find the number of moles of solute \(\left(\mathrm{SrBr}_{2}\right)\) in the solution, use the formula: Moles of solute = Volume of solution (in liters) × Molarity First, convert the volume to liters: \(600 \mathrm{~mL} = 0.600 \mathrm{~L}\) Now, multiply the volume in liters by the molarity of the solution: Moles of solute = \(0.600 \mathrm{~L} × 0.250 \mathrm{M}\) Moles of solute = \(0.150 \,\text{moles of}\, \mathrm{SrBr}_{2}\)
02

Solve for (b) using mass and molality

To find the number of moles of solute \(\left(\mathrm{KCl}\right)\) in the solution, use the formula: Moles of solute = Molality × Mass of solvent (in kilograms) First, convert the mass of solvent to kilograms: \(86.4 \mathrm{~g} = 0.0864 \mathrm{~kg}\) Now, multiply the mass of solvent in kilograms by the molality of the solution: Moles of solute = \(0.180 \mathrm{~m} × 0.0864 \mathrm{~kg}\) Moles of solute = \(0.0155 \,\text{moles of}\, \mathrm{KCl}\)
03

Solve for (c) using mass and mass percentage

To find the number of moles of solute \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in the solution, use the following steps: 1. Calculate the mass of glucose in the solution. 2. Calculate the moles of glucose using the molar mass of glucose. First, find the mass of glucose in the solution: Mass of glucose = \(6.45 \% × 124.0 \mathrm{~g}\) Mass of glucose = \(0.0645 × 124.0 \mathrm{~g} = 8.00 \mathrm{~g}\) Now, find the molar mass of glucose: Molar mass of glucose = \(6 × 12.01 \,\text{g/mol (C)} + 12 × 1.01 \,\text{g/mol (H)} + 6 × 16.00 \,\text{g/mol (O)} = 180.18 \,\text{g/mol}\) Finally, calculate the moles of glucose using the mass of glucose and its molar mass: Moles of solute = \(\frac{8.00 \mathrm{~g}}{180.18 \,\text{g/mol}}\) Moles of solute = \(0.0444 \,\text{moles of}\, \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) In summary, we have: (a) 0.150 moles of SrBr2 (b) 0.0155 moles of KCl (c) 0.0444 moles of C6H12O6

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain why pressure affects the solubility of \(\mathrm{O}_{2}\) in water, but not the solubility of \(\mathrm{NaCl}\) in water.

Acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a \(1.80 \mathrm{M}\) LiBr solution in acetonitrile is \(0.826 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the concentration of the solution in (a) molality, (b) mole fraction of LiBr, (c) mass percentage of \(\mathrm{CH}_{3} \mathrm{CN}\).

Why do ionic substances with higher lattice energies tend to be less soluble in water than those with lower lattice energies? [Section 13.1]

At \(20{ }^{\circ} \mathrm{C}\) the vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 75 torr, and that of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is 22 torr. Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fractions of a solution that has a vapor pressure of 35 torr at \(20^{\circ} \mathrm{C} ?\) (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?

Suppose you had a balloon made of some highly flexible semipermeable membrane. The balloon is filled com: pletely with a \(0.2 \mathrm{M}\) solution of some solute and is submerged in a \(0.1 \mathrm{M}\) solution of the same solute: Initially, the volume of solution in the balloon is \(0.25 \mathrm{~L} .\) Assuming the volume outside the semipermeable membrane is large, as the illustration shows, what would you expect for the solution volume inside the balloon once the system has come to equilibrium through osmosis? |Section 13.5]

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.