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Chapter 4: Problem 45

Water is evaporated from \(125 \mathrm{mL}\) of \(0.198 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}\) solution until the volume becomes \(105 \mathrm{mL}\). What is the molarity of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the remaining solution?

Short Answer

Expert verified
The molarity of the remaining solution is \(0.236 \mathrm{M}\)

Step by step solution

01

Convert initial volume to litres

The initial volume is given in milliliters, to use it in the calculations, it should be converted to liters. We know that 1L = 1000mL. So, the initial volume \(V_{1}\) in litres is \(125 \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.125 \mathrm{L}\)
02

Convert final volume to litres

Similarly, convert the final volume \(V_{2}\) to liters. So, \(105 \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.105 \mathrm{L}\)
03

Use Molarity Conservation Principle

Since the amount of solute remains the same before and after evaporation of water, we can use the molarity conservation equation \(C_{1}V_{1} = C_{2}V_{2}\). Setting up the equation \((0.198 M)(0.125 L) = C_{2}(0.105 L)\)
04

Solve for final molarity

Rearranging the equation for \(C_{2}\), we get \(C_{2} = \frac{(0.198 \mathrm{M})(0.125 \mathrm{L})}{0.105 \mathrm{L}}\)
05

Calculate final molarity

On calculation, \(C_{2} = 0.236 \mathrm{M}\), which represents the molarity of \(K_2SO_4\) in the remaining solution after evaporation

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Most popular questions from this chapter

A reaction mixture contains \(1.0 \mathrm{mol} \mathrm{CaCN}_{2}\) (calcium cyanamide) and \(1.0 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\). The maximum number of moles of \(\mathrm{NH}_{3}\) produced is (a) \(3.0 ;\) (b) 2.0 (c) between 1.0 and 2.0; (d) less than 1.0. $$\mathrm{CaCN}_{2}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{CaCO}_{3}+2 \mathrm{NH}_{3}(\mathrm{g})$$

For a specific reaction, ammonium dichromate is the only reactant and chromium(III) oxide and water are two of the three products. What is the third product and how many grams of this product are produced per kilogram of ammonium dichromate decomposed?

Titanium tetrachloride, \(\mathrm{TiCl}_{4}\) is prepared by the reaction below. $$\begin{aligned} &3 \mathrm{TiO}_{2}(\mathrm{s})+4 \mathrm{C}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{TiCl}_{4}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{CO}(\mathrm{g}) \end{aligned}$$ What is the maximum mass of \(\mathrm{TiCl}_{4}\) that can be obtained from \(35 \mathrm{g} \mathrm{TiO}_{2^{\prime}} 45 \mathrm{g} \mathrm{Cl}_{2^{\prime}}\) and \(11 \mathrm{g} \mathrm{C} ?\)

How many moles of \(\mathrm{NO}(\mathrm{g})\) can be produced in the reaction of \(3.00 \mathrm{mol} \mathrm{NH}_{3}(\mathrm{g})\) and \(4.00 \mathrm{mol} \mathrm{O}_{2}(\mathrm{g}) ?\) $$ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

Explain the important distinctions between (a) chemical formula and chemical equation; (b) stoichiometric coefficient and stoichiometric factor; (c) solute and solvent; (d) actual yield and percent yield; (e) consecutive and simultaneous reactions.
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