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Chapter 11: Problem 14

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 L of a 0.100 M sodium oxalate \(\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) solution, 3.35 g of sodium oxalate is needed.

Step by step solution

01

Calculate the number of moles of sodium oxalate required

To calculate the number of moles of sodium oxalate required for a 0.100 M solution, we can use the formula: Moles (n) = Molarity (M) x Volume (L) The molarity is given as 0.100 M, and the volume is given as 0.250 L. Plug in these values and solve for the number of moles of sodium oxalate: n = 0.100 M x 0.250 L = 0.025 moles
02

Calculate the molar mass of sodium oxalate

To find the mass of sodium oxalate needed, we need to know the molar mass of sodium oxalate. Sodium oxalate is Na2C2O4, so its molar mass can be calculated by summing the molar masses of the individual elements: Molar mass of Na2C2O4 = 2(Mass of Na) + 2(Mass of C) + 4(Mass of O) Using the atomic masses from the periodic table: Molar mass of Na2C2O4 = 2(22.989 g/mol) + 2(12.011 g/mol) + 4(16.000 g/mol) = 133.998 g/mol
03

Find the mass of sodium oxalate needed

Now that we have both the number of moles required and the molar mass of sodium oxalate, we can find the mass of sodium oxalate needed to prepare the solution. Mass = Moles x Molar mass Plug in the number of moles (0.025 moles) and the molar mass (133.998 g/mol) of sodium oxalate: Mass = 0.025 moles x 133.998 g/mol = 3.34995 g
04

Round to an appropriate number of significant figures

Since the volume and molarity in the question are given to three significant figures, we'll round our answer to three significant figures as well: Mass of sodium oxalate needed = 3.35 g To prepare 0.250 L of a 0.100 M sodium oxalate solution, 3.35 g of sodium oxalate is needed.

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