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Chapter 11: Problem 12

What mass of sodium oxalate \(\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is needed to prepare \(0.250 \mathrm{~L}\) of a \(0.100 M\) solution?

Short Answer

Expert verified
To prepare \(0.250 \mathrm{~L}\) of a \(0.100 \mathrm{~M}\) sodium oxalate \(\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) solution, you will need \(3.35 \mathrm{~g}\) of sodium oxalate.

Step by step solution

01

Calculate the number of moles of Na2C2O4 needed

First, we need to determine the number of moles of sodium oxalate required for the solution. We can use the formula: moles = molarity × volume where moles represent the amount of solute (Na2C2O4), molarity refers to the concentration of the solution (0.100 M), and volume is the volume of the solution (0.250 L). So we can plug the given values: moles = (0.100 M) × (0.250 L)
02

Calculate the moles of Na2C2O4

Now we can calculate the moles of sodium oxalate required: moles = (0.1 moles/L) × (0.25 L) moles = 0.025 moles of Na2C2O4
03

Determine the molar mass of Na2C2O4

To find the mass of sodium oxalate, we will need to determine its molar mass. The molar mass of a compound is the sum of the atomic masses of each element in the compound. In the case of Na2C2O4, we have: - 2 Na atoms with an atomic mass of 22.99 g/mol - 2 C atoms with an atomic mass of 12.01 g/mol - 4 O atoms with an atomic mass of 16.00 g/mol Molar mass of Na2C2O4 = (2 × 22.99) + (2 × 12.01) + (4 × 16.00)
04

Calculate the molar mass of Na2C2O4

Based on the atomic masses, we can now calculate the molar mass of sodium oxalate: Molar mass of Na2C2O4 = (2 × 22.99) + (2 × 12.01) + (4 × 16.00) = 45.98 + 24.02 + 64.00 = 134.00 g/mol
05

Calculate the mass of Na2C2O4 required

Now we can determine the mass of sodium oxalate needed by multiplying the moles we calculated in Step 2 by the molar mass we found in Step 4: mass = moles × molar mass mass = (0.025 moles) × (134.00 g/mol)
06

Find the mass of Na2C2O4 needed

Finally, we can find the mass of sodium oxalate required to prepare the solution: mass = (0.025 moles) × (134.00 g/mol) = 3.35 g So, 3.35 grams of sodium oxalate (Na2C2O4) is needed to prepare 0.250 L of a 0.100 M solution.

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Most popular questions from this chapter

An aqueous solution of \(10.00 \mathrm{~g}\) of catalase, an enzyme found in the liver, has a volume of \(1.00 \mathrm{~L}\) at \(27^{\circ} \mathrm{C}\). The solution's osmotic pressure at \(27^{\circ} \mathrm{C}\) is found to be \(0.745\) torr. Calculate the molar mass of catalase.

For an acid or a base, when is the normality of a solution equal to the molarity of the solution and when are the two concentration units different?

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