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Chapter 11: Problem 12

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

Short Answer

Expert verified
To prepare 0.250 L of a 0.100-M solution of sodium oxalate (Na₂C₂O₄), you need 3.35 g of sodium oxalate.

Step by step solution

01

Determine the moles of sodium oxalate required to prepare the solution

We are given the volume of the solution (0.250 L) and the molarity (0.100 M). Use the formula to find the moles of sodium oxalate: Moles of sodium oxalate = Molarity × Volume $$ = 0.100\,M\times0.250\,L $$ $$ = 0.025\, moles $$
02

Calculate the molar mass of sodium oxalate

To find the mass of sodium oxalate, we need to calculate the molar mass of sodium oxalate (Na₂C₂O₄). The molar mass can be calculated by adding the molar masses of the elements present in it. Molar mass of Na₂C₂O₄ = 2 × (Molar mass of Na) + 2 × (Molar mass of C) + 4 × (Molar mass of O) From the periodic table, we can find the molar masses of the elements: Na = 22.99 g/mol, C = 12.01 g/mol, and O = 16.00 g/mol. Molar mass of Na₂C₂O₄ = 2(22.99 g/mol) + 2(12.01 g/mol) + 4(16.00 g/mol) $$ = 45.98\,g/mol + 24.02\,g/mol + 64.00\,g/mol $$ $$ = 134.00\,g/mol $$
03

Find the mass of sodium oxalate required

Now that we have the moles of sodium oxalate and its molar mass, we can find the mass of sodium oxalate required to prepare the solution: Mass of sodium oxalate = moles × molar mass $$ = 0.025\,moles \times 134.00\,g/mol $$ $$ = 3.35\,g $$ Thus, 3.35 g of sodium oxalate is needed to prepare 0.250 L of a 0.100-M solution.

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Most popular questions from this chapter

If the fluid inside a tree is about \(0.1 M\) more concentrated in solute than the groundwater that bathes the roots, how highwill a column of fluid rise in the tree at \(25^{\circ} \mathrm{C}\) ? Assume that the density of the fluid is \(1.0 \mathrm{~g} / \mathrm{cm}^{3}\). (The density of mercury is \(\left.13.6 \mathrm{~g} / \mathrm{cm}^{3} .\right)\)

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