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Chapter 4: Problem 67

Calculate (a) the number of grams of solute in \(0.250 \mathrm{~L}\) of \(0.175 \mathrm{M} \mathrm{KBr},\) (b) the molar concentration of a solution containing \(14.75 \mathrm{~g}\) of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(1.375 \mathrm{~L},\) (c) the volume of \(1.50 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) in milliliters that contains \(2.50 \mathrm{~g}\) of solute.

Short Answer

Expert verified
(a) 5.21 g of KBr (b) 0.0654 M Ca(NO\(_3\))\(_2\) (c) 10.2 mL of 1.50 M Na\(_3\)PO\(_4\)

Step by step solution

01

Calculate moles of solute in KBr solution

To determine the moles of solute in the KBr solution, use the formula: moles of solute = molarity × volume moles of \(KBr = 0.175\,M × 0.250\,L = 0.04375\,mol\)
02

Calculate grams of solute in KBr solution

Now, we'll convert the moles of solute to grams using the molar mass of KBr. Molar mass of KBr = 39.1 g/mol (for K) + 79.9 g/mol (for Br) = 119 g/mol Grams of KBr = moles × molar mass = \(0.04375\,mol × 119\,g/mol = 5.21\,g\) So, there are 5.21 grams of KBr solute.
03

Calculate moles of solute in Ca(NO\(_3\))\(_2\) solution

First, find the moles of solute using the mass and molar mass of Ca(NO\(_3\))\(_2\). Molar mass of Ca(NO\(_3\))\(_2\) = 40.08 g/mol (for Ca) + 2 × (14.01 g/mol (for N) + 3 × 16.00 g/mol (for O)) = 164.10 g/mol Moles of Ca(NO\(_3\))\(_2\) = mass / molar mass = \(14.75\,g / 164.10\,g/mol = 0.0899\,mol\)
04

Calculate the molar concentration of the Ca(NO\(_3\))\(_2\) solution

Now, we'll calculate the molar concentration using the formula: molarity = moles of solute / volume Molarity of Ca(NO\(_3\))\(_2\) = \(0.0899\,mol / 1.375\,L = 0.0654\,M\) So, the molar concentration of the Ca(NO\(_3\))\(_2\) solution is 0.0654 M.
05

Calculate moles of solute in Na\(_3\)PO\(_4\) solution

First, find the moles of solute using the mass and molar mass of Na\(_3\)PO\(_4\). Molar mass of Na\(_3\)PO\(_4\) = 3 × 22.99 g/mol (for Na) + 30.97 g/mol (for P) + 4 × 16.00 g/mol (for O) = 163.94 g/mol Moles of Na\(_3\)PO\(_4\) = mass / molar mass = \(2.50\,g / 163.94\,g/mol = 0.0153\,mol\)
06

Calculate the volume of Na\(_3\)PO\(_4\) solution

Now, find the volume in liters using the formula: volume = moles of solute / molarity Volume of Na\(_3\)PO\(_4\) = \(0.0153\,mol / 1.50\,M = 0.0102\,L\)
07

Convert the volume to milliliters

Finally, convert the volume from liters to milliliters. Volume of Na\(_3\)PO\(_4\) = \(0.0102\,L × 1000\,mL/L = 10.2\,mL\) So, the volume of 1.50 M Na\(_3\)PO\(_4\) in milliliters that contains 2.50 g of solute is 10.2 mL.

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Most popular questions from this chapter

(a) A strontium hydroxide solution is prepared by dissolving \(10.45 \mathrm{~g}\) of \(\mathrm{Sr}(\mathrm{OH})_{2}\) in water to make \(50.00 \mathrm{~mL}\) of solution. What is the molarity of this solution? (b) Next the strontium hydroxide solution prepared in part (a) is used to titrate a nitric acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between strontium hydroxide and nitric acid solutions. (c) If \(23.9 \mathrm{~mL}\) of the strontium hydroxide solution was needed to neutralize a \(31.5 \mathrm{~mL}\) aliquot of the nitric acid solution, what is the concentration (molarity) of the acid?

(a) How many milliliters of a stock solution of \(6.0 \mathrm{M} \mathrm{HNO}_{3}\) would you have to use to prepare \(110 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HNO}_{3} ?\) (b) If you dilute \(10.0 \mathrm{~mL}\) of the stock solution to a final volume of \(0.250 \mathrm{~L}\), what will be the concentration of the diluted solution?

A solution of \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH}\) is mixed with a solution of \(200.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{NiSO}_{4}\). (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

(a) Use the following reactions to prepare an activity series for the halogens: $$ \begin{aligned} \mathrm{Br}_{2}(a q)+2 \mathrm{NaI}(a q) & \longrightarrow 2 \mathrm{NaBr}(a q)+\mathrm{I}_{2}(a q) \\ \mathrm{Cl}_{2}(a q)+2 \mathrm{NaBr}(a q) & \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(a q) \end{aligned} $$ (b) Relate the positions of the halogens in the periodic table with their locations in this activity series. (c) Predict whether a reaction occurs when the following reagents are mixed: \(\mathrm{Cl}_{2}(a q)\) and \(\mathrm{KI}(a q) ; \operatorname{Br}_{2}(a q)\) and \(\operatorname{LiCl}(a q)\)

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-},(\mathrm{b}) \mathrm{NO}_{3}^{-},(\mathrm{c}) \mathrm{NH}_{4}^{+},(\mathrm{d}) \mathrm{S}^{2-},\) (e) \(\mathrm{SO}_{4}^{2-}\). Explain briefly. [Section \(\left.4.2\right]\)
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