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Chapter 13: Problem 11

A certain brine has \(3.87 \%\) NaCl by mass. A \(75.0 \mathrm{mL}\) sample weighs 76.9 g. How many liters of this solution should be evaporated to dryness to obtain \(725 \mathrm{kg}\) \(\mathrm{NaCl} \)?

Short Answer

Expert verified
You would need to evaporate approximately 18614.67 liters of the solution.

Step by step solution

01

Find the Mass of NaCl in the Given Sample

From the problem, we know that the brine is \(3.87\%\) NaCl by mass. Since we are given the mass of the sample (76.9 g), we can find the mass of NaCl in the sample by multiplying it by \(3.87/100\), and get \(m_{NaCl} = 76.9 g * 3.87/100 = 2.979 g\).\nThis says there are 2.979 g of NaCl in every 76.9 g of solution.
02

Convert the Desired Mass of NaCl to Grams

We're asked to find the volume of solution necessary to obtain 725 kg of NaCl. We need to convert this mass from kilograms to grams, since our previous calculations were done in grams. There are 1000 g in 1 kg, so \(m'_{NaCl} = 725 kg * 1000 = 725000 g\).
03

Calculate the Required Volume of the Solution

We know there are 2.979 g of NaCl in every 76.9 g (or mL, since the density of the solution is roughly 1 g/mL) of solution. We can set up a proportion to find out how much solution we would need to get 725000 g of NaCl.\nLet \(x\) be the required volume, then we can write as:\[ \frac{2.979 g}{76.9 mL} = \frac{725000 g}{x} \]Solving for \(x\), we get \(x = \frac{76.9 mL * 725000 g}{2.979 g} = 18614666.73 mL = 18614.67 L\].
04

Convert the Volume to Liters

As the final answer should be given in liters instead of milliliters, we need to convert \(x\) from mL to L. There are 1000 mL in 1 L, so we get 18614.67 L.

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