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Chapter 4: Problem 41

Analysis shows that a sample of \(\mathrm{H}_{2} \mathrm{X}\) (MM \(=100.0 \mathrm{~g} / \mathrm{mol}\) ) reacts completely with \(330.0 \mathrm{~mL}\) of \(0.2000 \mathrm{M} \mathrm{KOH}\). $$ \mathrm{H}_{2} \mathrm{X}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{X}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O} $$ What is the volume of the sample? (Density of \(\mathrm{H}_{2} \mathrm{X}=1.200 \mathrm{~g} / \mathrm{mL}\).)

Short Answer

Expert verified
Answer: The volume of the sample of H2X is 2.75 mL.

Step by step solution

01

Use volume and molarity of KOH to get moles of KOH

Using the volume and molarity of \(\mathrm{KOH}\), we can calculate the moles of \(\mathrm{KOH}\). Molarity is given by the formula \(\text{Molarity}(\mathrm{M})=\frac{\text{moles of solute}}{\text{volume of solution in liters}}\). Moles of KOH = molarity × volume $$ \text{moles of KOH}=0.2000\,\text{M} \times 0.330\,\text{L}=0.0660\,\text{moles} $$
02

Use the stoichiometry of the balanced chemical equation to find moles of H2X

In the balanced chemical equation, we see that the ratio of \(\mathrm{H}_{2}\mathrm{X}\) to \(\mathrm{KOH}\) is \(1:2\). Hence, we can find the moles of \(\mathrm{H}_{2}\mathrm{X}\) using stoichiometric coefficients. Moles of H2X = (Moles of KOH)/2.0 $$ \text{moles of H2X}=\frac{0.0660\,\text{moles}}{2} = 0.0330\,\text{moles} $$
03

Convert moles of H2X to mass using molar mass

We are given that the molar mass of \(\mathrm{H}_{2}\mathrm{X}\) is 100 g/mol. Using this information, we can convert the moles of \(\mathrm{H}_{2}\mathrm{X}\) to grams: mass = moles × molar mass $$ \text{mass of H2X} = 0.0330\,\text{moles} \times 100.0\,\text{g/mol} = 3.30\,\text{g} $$
04

Calculate the volume of the sample using density

We have the mass of the sample, and we are also given the density of \(\mathrm{H}_{2}\mathrm{X}\). Using the formula for density (\(\text{density} = \frac{\text{mass}}{\text{volume}}\)), we can calculate the volume of the sample. volume = mass/density $$ \text{volume of H2X} = \frac{3.30\,\text{g}}{1.200\,\text{g/mL}} = 2.75\,\text{mL} $$
05

Final answer

The volume of the sample of \(\mathrm{H}_{2}\mathrm{X}\) is \(2.75\,\text{mL}\).

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