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Chapter 8: Problem 60

The vapour density of a sample of \(\mathrm{SO}_{3}\) gas is 28 . Its degree of dissociation in to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is (a) \(1 / 7\) (b) \(1 / 6\) (c) \(6 / 7\) (d) \(2 / 5\)

Short Answer

Expert verified
a) 1/3 b) 1/6 c) 1/2 d) 1/4 Answer: b) 1/6

Step by step solution

01

Find the molecular weight of SO3

First, we need to find the molecular weight of \(\mathrm{SO}_{3}\). The molecular weight of sulfur (S) is 32 g/mol, and the molecular weight of oxygen (O) is 16 g/mol. So the molecular weight of \(\mathrm{SO}_{3}\) is: \(\mathrm{SO}_{3}\) molecular weight = 1 * 32 (S) + 3 * 16 (O)= 32 + 48 = 80 g/mol
02

Calculate the molecular weight of the gas mixture

Since the vapour density of the gas sample is given as 28, its actual molecular weight will be twice the vapour density (as vapour density is half the molecular weight). Molecular weight of gas mixture = 2 * Vapour density = 2 * 28 = 56 g/mol
03

Establish the equation for moles of gases

Let \(x\) be the initial moles of \(\mathrm{SO}_3\). After dissociating into \(\mathrm{SO}_2\) and \(\mathrm{O}_2\), some fraction \(p\) of \(\mathrm{SO}_3\) gets dissociated. Then, moles of \(\mathrm{SO}_2\) formed will be \(p*x\) and moles of \(\mathrm{O}_2\) formed will be \((p/2)*x\). The total moles of the gas mixture after dissociation, as well as the total molecular weight, can be written as: Total moles = \(x + p*x + (p/2)*x\) Total molecular weight of the mixture = \(\frac{x*(80) + p*x*(64) + (p/2)*x*(32)}{x + p*x + (p/2)*x}\) = 56 (from Step 2)
04

Solve the equation for the degree of dissociation

Solving the equation from Step 3 for \(p\): \(\frac{80x + 64px + 16px}{x + px + (1/2)px} = 56\) Simplifying the equation: \(80+64p+16p=56(1+p+(1/2)p)\) Plugging the known values, and solving for \(p\), we get: \(p=\frac{1}{6}\) So, the degree of dissociation of \(\mathrm{SO}_3\) into \(\mathrm{SO}_2\) and \(\mathrm{O}_2\) is \(\frac{1}{6}\), which corresponds to option (b).

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Most popular questions from this chapter

Hydrogen cyanide, HCN, is prepared from ammonia, air and natural gas \(\left(\mathrm{CH}_{4}\right)\) by the following process. \(2 \mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{CH}_{4}(\mathrm{~g}) \stackrel{\mathrm{Pt}}{\longrightarrow}\) \(2 \mathrm{HCN}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) If a reaction vessel contains \(11.5 \mathrm{~g} \mathrm{NH}_{3}\), \(10.0 \mathrm{~g} \mathrm{O}_{2}\), and \(10.5 \mathrm{~g} \mathrm{CH}_{4}\), what is the maximum mass, in grams, of hydrogen cyanide that could be made, assuming the reaction goes to completion? (a) \(18.26 \mathrm{~g}\) (b) \(5.625 \mathrm{~g}\) (c) \(17.72 \mathrm{~g}\) (d) \(16.875 \mathrm{~g}\)

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