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Chapter 4: Problem 59

Calculate the volume in \(\mathrm{mL}\) of a solution required to provide the following: (a) \(2.14 \mathrm{~g}\) of sodium chloride from a \(0.270 \mathrm{M}\) solution, (b) \(4.30 \mathrm{~g}\) of ethanol from a \(1.50 M\) solution, (c) 0.85 g of acetic acid ( \(\left.\mathrm{CH}_{3} \mathrm{COOH}\right)\) from a \(0.30 \mathrm{M}\) solution.

Short Answer

Expert verified
The volumes required for each solution are as follows: (a) 135.5 mL of the sodium chloride solution, (b) 62.2 mL of the ethanol solution, and (c) 47.2 mL of the acetic acid solution.

Step by step solution

01

Conversion into Moles

Firstly, we need to convert the given masses into moles. The number of moles (n) can be found by dividing the given mass (m) of each substance by its molar mass (Mm). For (a) sodium chloride (NaCl), the molar mass is approximately 58.44 g/mol. So, \(n_{NaCl} = \frac{m_{NaCl}}{Mm_{NaCl}} = \frac{2.14 g}{58.44 g/mol} = 0.0366 mol\)For (b) ethanol (C2H5OH), the molar mass is approximately 46.07 g/mol. So, \(n_{Ethanol} = \frac{m_{Ethanol}}{Mm_{Ethanol}} = \frac{4.30 g}{46.07 g/mol} = 0.0933 mol\)For (c) acetic acid (CH3COOH), the molar mass is approximately 60.05 g/mol. So, \(n_{Acetic Acid} = \frac{m_{Acetic Acid}}{Mm_{Acetic Acid}} = \frac{0.85 g}{60.05 g/mol} = 0.0142 mol\)
02

Applying Molarity Formula

Secondly, we use the molarity formula to calculate the volume required for each solution. The molarity formula can be rearranged to find the volume (V) as follows: \(V=\frac{n}{M}\), where V is the volume in liters. To get the volume in milliliters (mL), we multiply by 1000. For (a) the sodium chloride solution: \(V_{NaCl} = \frac{n_{NaCl}}{M_{NaCl}} * 1000 = \frac{0.0366 mol}{0.270 M} * 1000 = 135.5 mL\)For (b) the ethanol solution: \(V_{Ethanol} = \frac{n_{Ethanol}}{M_{Ethanol}} * 1000 = \frac{0.0933 mol}{1.50 M} * 1000 = 62.2 mL\)For (c) the acetic acid solution: \(V_{Acetic Acid} = \frac{n_{Acetic Acid}}{M_{Acetic Acid}} * 1000 = \frac{0.0142 mol}{0.30 M} * 1000 = 47.2 mL\)

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