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Chapter 13: Problem 19

An aqueous solution is \(6.00 \%\) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) by mass, with \(d=0.988 \mathrm{g} / \mathrm{mL} .\) What is the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in this solution?

Short Answer

Expert verified
The molarity of the Methanol in the solution is 1.85 M.

Step by step solution

01

Find the mass of the solution

First, the total mass of the solution is needed. Given that the density (d) is 0.988 g/mL, and considering 1L of solution (since molarity is moles/Liter), the total mass can be found by converting the volume in mL to L, for 1L, it's 1000mL. Thus, the mass of the solution = d * V = 0.988 g/mL * 1000 mL = 988 g.
02

Find the mass of Methanol

Next, find the mass of the Methanol (solute). From the problem, we know that the solution is 6.00 % by mass Methanol. So, Methanol will be 6.00 % of the total mass of the solution found in the previous step. Therefore, Mass of Methanol = 6.00/100 * Total mass of solution = 0.0600 * 988 g = 59.28 g.
03

Find the number of moles

To find molarity, the number of moles is needed. Therefore, the number of moles of the Methanol needs to be calculated using its molar mass. The molar mass of Methanol (CH3OH) is approximately 32.04 g/mole. So, Moles of Methanol = Mass/Molar mass = 59.28 g / 32.04 g/mol = 1.85 moles.
04

Calculate Molarity

Finally, we can calculate the molarity (M), which is the number of moles per Liter of solution. From earlier steps, we've determined that there are 1.85 moles of Methanol in 1L of solution. So the molarity (M) is just this value M = 1.85 moles / 1 L = 1.85 M.

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Most popular questions from this chapter

Which of the following ions has the greater charge density? (a) \(\mathrm{Na}^{+} ;\) (b) \(\mathrm{F}^{-} ;\) (c) \(\mathrm{K}^{+} ;\) (d) \(\mathrm{Cl}^{-}\).

A saturated aqueous solution of \(\mathrm{NaBr}\) at \(20^{\circ} \mathrm{C}\) contains \(116 \mathrm{g} \mathrm{NaBr} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). Express this composition in the more conventional percent by mass, that is, as grams of NaBr per 100 grams of solution.

A typical commercial grade aqueous phosphoric acid is \(75 \% \mathrm{H}_{3} \mathrm{PO}_{4}\) by mass and has a density of \(1.57 \mathrm{g} / \mathrm{mL}\) What is the molarity of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in this solution?

The most likely of the following mixtures to be an ideal solution is (a) \(\mathrm{NaCl}-\mathrm{H}_{2} \mathrm{O} ;\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}-\mathrm{C}_{6} \mathrm{H}_{6}\) (c) \(\mathrm{C}_{7} \mathrm{H}_{16}-\mathrm{H}_{2} \mathrm{O} ;\) (d) \(\mathrm{C}_{7} \mathrm{H}_{16}-\mathrm{C}_{8} \mathrm{H}_{18}\)

At \(25^{\circ} \mathrm{C}\) and under an \(\mathrm{O}_{2}(\mathrm{g})\) pressure of \(1 \mathrm{atm},\) the solubility of \(\mathrm{O}_{2}(\mathrm{g})\) in water is \(28.31 \mathrm{mL} / 1.00 \mathrm{L} \mathrm{H}_{2} \mathrm{O}\) At \(25^{\circ} \mathrm{C}\) and under an \(\mathrm{N}_{2}(\mathrm{g})\) pressure of \(1 \mathrm{atm},\) the solubility of \(\mathrm{N}_{2}(\mathrm{g})\) in water is \(14.34 \mathrm{mL} / 1.00 \mathrm{L} \mathrm{H}_{2} \mathrm{O}\) The composition of the atmosphere is \(78.08 \% \mathrm{N}_{2}\) and \(20.95 \% \mathrm{O}_{2},\) by volume. What is the composition of air dissolved in water expressed as volume percents of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2} ?\)
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