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Chapter 4: Problem 33

What mass of \(\mathrm{NaOH}\) is contained in \(250.0 \mathrm{~mL}\) of a \(0.400 \mathrm{M}\) sodium hydroxide solution?

Short Answer

Expert verified
The mass of \(\mathrm{NaOH}\) contained in \(250.0 \mathrm{~mL}\) of a \(0.400\mathrm{M}\) sodium hydroxide solution is 4.00 g.

Step by step solution

01

Convert mL of solution to L

To convert from mL to L, we'll divide the volume of the solution (250.0 mL) by 1000 since there are 1000 mL in 1 L: Volume in L = 250.0 mL 梅 1000 = 0.250 L
02

Calculate the number of moles of NaOH

To find the number of moles of NaOH, we'll multiply the volume of the solution (in L) by the given concentration: Number of moles of NaOH = (volume of solution in L) 脳 (concentration) Number of moles of NaOH = (0.250 L) 脳 (0.400 mol/L) = 0.100 mol
03

Find the molar mass of NaOH

To calculate the mass of NaOH, we need the molar mass of the solute. The formula for sodium hydroxide is NaOH, and we can find the molar mass using the periodic table: Molar mass of Na (sodium) = 22.99 g/mol Molar mass of O (oxygen) = 16.00 g/mol Molar mass of H (hydrogen) = 1.008 g/mol Molar mass of NaOH = (22.99 + 16.00 + 1.008) g/mol = 40.00 g/mol
04

Calculate the mass of NaOH

Now we can calculate the mass of NaOH by multiplying the number of moles with the molar mass: mass of NaOH = (number of moles of NaOH) 脳 (molar mass of NaOH) mass of NaOH = (0.100 mol) 脳 (40.00 g/mol) = 4.00 g The mass of \(\mathrm{NaOH}\) contained in \(250.0 \mathrm{~mL}\) of a \(0.400\mathrm{M}\) sodium hydroxide solution is 4.00 g.

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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.

You made \(100.0 \mathrm{~mL}\) of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only \(80.0 \mathrm{~mL}\) left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take \(2.00 \mathrm{~mL}\) of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of \(3.407 \mathrm{~g}\). What was the concentration of the original lead(II) nitrate solution?

Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations. copper(II) sulfate \((a q)+\operatorname{iron}(s)\) $$ \operatorname{copper}(s)+\text { iron(II) sulfate }(a q) $$ \(\operatorname{copper}(\) II \()\) sulfate \((a q)+\operatorname{iron}(s)\) $$ \operatorname{copper}(s)+\text { iron(III) sulfate }(a q) $$ You place \(87.7 \mathrm{~mL}\) of a \(0.500 M\) solution of copper(II) sulfate in a beaker. You then add \(2.00 \mathrm{~g}\) of iron filings to the copper(II) sulfate solution. After one of the above reactions occurs, you isolate \(2.27 \mathrm{~g}\) of copper. Which equation above describes the reaction that occurred? Support your answer.

Suppose \(50.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{CoCl}_{2}\) solution is added to \(25.0 \mathrm{~mL}\) of \(0.350 \mathrm{M} \mathrm{NiCl}_{2}\) solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume that the volumes are additive.

Show how each of the following strong electrolytes "breaks up" into its component ions upon dissolving in water by drawing molecular-level pictures. a. \(\mathrm{NaBr}\) f. \(\mathrm{FeSO}_{4}\) b. \(\mathrm{MgCl}_{2}\) g. \(\mathrm{KMnO}_{4}\) c. \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) h. \(\mathrm{HClO}_{4}\) d. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) i. \(\mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (ammonium acetate) e. \(\mathrm{NaOH}\)
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