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Chapter 4: Problem 100

A 230.-mL sample of a \(0.275 \mathrm{M} \mathrm{CaCl}_{2}\) solution is left on a hot plate overnight; the following morning, the solution is \(1.10 \mathrm{M}\). What volume of water evaporated from the \(0.275 \mathrm{M} \mathrm{CaCl}_{2}\) solution?

Short Answer

Expert verified
172.5 mL of water evaporated from the 0.275 M CaCl2 solution.

Step by step solution

01

Calculate the initial amount of solute in the solution.

First, calculate the initial amount (in moles) of the calcium chloride in the 230 mL solution by using the formula: moles = concentration x volume. Initial moles of CaCl2 = \( 0.275 \mathrm{M} \times 230 \mathrm{mL} \) Please note that it's essential to convert the volume from milliliters to liters: Initial moles of CaCl2 = \( 0.275 \mathrm{M} \times 0.230 \mathrm{L} \) Initial moles of CaCl2 = \( 0.06325 \mathrm{mol} \)
02

Calculate the final volume of the solution after water evaporation.

Since the amount of solute (calcium chloride) is conserved during the evaporation process, we can now calculate the final volume of the 1.10 M solution. We will use the formula: moles = concentration x volume. Rearranging it for volume: Final volume = moles / concentration Final volume = \( 0.06325 \mathrm{mol} \div 1.10 \mathrm{M} \) Final volume = \( 0.0575 \mathrm{L} \)
03

Calculate the volume of water evaporated from the original solution.

Now, subtract the final volume from the initial volume to get the volume of water evaporated: Volume of water evaporated = Initial volume - Final volume Volume of water evaporated = \( 0.230 \mathrm{L} - 0.0575 \mathrm{L} \) Volume of water evaporated = \( 0.1725 \mathrm{L} \) Finally, convert the volume of water evaporated from liters to milliliters: Volume of water evaporated = \( 0.1725 \mathrm{L} \times 1000 \mathrm{mL/L} \) Volume of water evaporated = \( 172.5 \mathrm{mL} \) So, 172.5 mL of water evaporated from the 0.275 M CaCl2 solution.

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Most popular questions from this chapter

A \(100.0-\mathrm{mL}\) aliquot of \(0.200 M\) aqueous potassium hydroxide is mixed with \(100.0 \mathrm{~mL}\) of \(0.200 M\) aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .\) A \(1.45-\mathrm{g}\) sample of the mixture is dissolved in water and an excess of \(\mathrm{NaOH}\) is added, producing a precipitate of \(\mathrm{Al}(\mathrm{OH})_{3} .\) The precipitate is filtered. dried, and weighed. The mass of the precipitate is \(0.107 \mathrm{~g}\). What is the mass percent of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the sample?

What mass of solid \(\mathrm{AgBr}\) is produced when \(100.0 \mathrm{~mL}\) of \(0.150 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) is added to \(20.0 \mathrm{~mL}\) of \(1.00 M \mathrm{NaBr} ?\)

Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: \(100.0 \mathrm{~mL}\) of \(0.30 \mathrm{M} \mathrm{AlCl}_{3}, 50.0 \mathrm{~mL}\) of \(0.60 \mathrm{M} \mathrm{MgCl}_{2}\), or \(200.0 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{NaCl} ?\)

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