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Chapter 30: Problem 8

Which is more concentrated with respect to sodium ions: \(50.0 \mathrm{~g}\) of \(\mathrm{NaCl}\) in \(500.0 \mathrm{~mL}\) of solution or \(59.0 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in \(500.0 \mathrm{~mL}\) of solution?

Short Answer

Expert verified
Without calculating the exact values, we can ascertain that the solution with Na2SO4 will have a higher concentration of sodium ions because each mole of Na2SO4 produces two moles of sodium ions, unlike NaCl where each mole produces only one mole of sodium ions. Thus, the 59g of Na2SO4 in 500mL of the solution is more concentrated with respect to sodium ions.

Step by step solution

01

Calculate the Amount of Sodium Ions in Each Compound

Firstly, find the number of moles of each compound. For NaCl, it is composed of one sodium ion(Na) thus the amount of Na ions in the compound is equal to the number of moles of the compound. For Na2SO4, it is composed of two sodium ions(Na), so the amount of Na ions in the compound is twice the number of moles of the compound. The formula to calculate moles is: Moles = mass (g) / molecular weight (g/mol). The molecular weight of NaCl is 58.44 g/mol and that of Na2SO4 is 142.04 g/mol.
02

Calculate the Concentration of Sodium Ions in Each Solution

After finding the moles of sodium ions, we can calculate the concentration of Na ions in each solution. The volume of both solutions is 500 mL which is equal to 0.5 liter. Therefore, the formula to calculate the concentration for each solution is: concentration = Moles / Volume (L). Apply this formula to both compounds to find the concentration of Na ions.
03

Compare the Concentrations

Once the concentrations of sodium ions in each solution are known, they can be compared. The solution with the greater concentration of sodium ions will be the one that is more concentrated with respect to sodium ions.

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Most popular questions from this chapter

Draw a figure similar to the right side of Figure 1 for \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) in solution showing exactly 3 cations and the appropriate number of anions. (You can omit \(\mathrm{H}_{2} \mathrm{O}\) molecules for clarity. \()\)

a) Which atom(s) of the water molecule is closest to the cations in solution? Why? b) Which atom(s) of the water molecule is closest to the anions in solution? Why?

Verify that when \(10.0 \mathrm{~g}\) of sodium sulfate dissolves in water: a) there are \(7.04 \times 10^{-2}\) moles of sodium sulfate in the water. b) there are \(7.04 \times 10^{-2}\) moles of sulfate in the water. c) there are \(14.1 \times 10^{-2}\) moles of sodium in the water.

Which is more concentrated with respect to sodium ions, \(0.30 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) or \(0.50 \mathrm{M} \mathrm{NaCl} ?\)

When \(1.000\) mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is dissolved in water to make exactly one liter of solution: a) how many moles of sodium ions are found in the solution? b) how many moles of sulfate ions are found in the solution? c) what is the molarity of sodium ions in the solution? d) what is the molarity of sulfate ions in the solution?
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