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Chapter 2: Problem 98

One oxide of rubidium has \(0.187 \mathrm{g}\) O per gram of Rb. A possible O:Rb mass ratio for a second oxide of rubidium is (a) \(16: 85.5 ;\) (b) \(8: 42.7 ;\) (c) \(1: 2.674 ;\) (d) any of these.

Short Answer

Expert verified
The possible O:Rb mass ratio for a second oxide of rubidium is (a) 16: 85.5

Step by step solution

01

Reinterpretation of the Given Ratio

An important starting point is the relationship that \(0.187 \mathrm{g}\) of O combine with \(1 \mathrm{g}\) of Rb in the first oxide. This implies that every gram of rubidium carries with it \(0.187 \mathrm{g}\) of oxygen. It is essential to realize that in terms of atomic amounts it means one mole of Rb is combining with a certain number of moles of O.
02

Calculation of Moles in the Established Ratio

To use this ratio, convert the mass of oxygen to moles, one mole of O has a mass of \(16 \mathrm{g}\). Thus \(0.187 g\) of O corresponds to: \(0.187 / 16 = 0.0117 \, \mathrm{moles}\). So, in the first oxide, 1 mole of Rb is combining with \(0.0117 \, \mathrm{moles}\) of O.
03

Examination of the Second Oxide Ratio

In the second oxide, the O:Rb mass ratio must again result in \(1 \, \mathrm{mole}\) of Rb combining with some integer multiple of \(0.0117 \, \mathrm{moles}\) of O. As molar mass of Rb is \(85.5 \, \mathrm{g}\), option (a) \(16: 85.5\) means that for \(85.5 \, \mathrm{g}\) of Rb (1 mole), \(16 \, \mathrm{g}\) or \(1 \, \mathrm{mole}\) of O is present. Thus, the mole ratio of O to Rb is equal to \(1/0.0117 = 85.0\). So, option (a) seems to be a valid ratio.
04

Verification of the Other Options

However, options (b) and (c) do not result in the mole of O being an integer multiple of \(0.0117 \) moles, thus these are not valid ratios for the second oxide of rubidium.
05

Final Answer

Therefore, only option (a) is a valid O:Rb mass ratio for a second oxide of rubidium.

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Most popular questions from this chapter

How many atoms are present in a \(75.0 \mathrm{cm}^{3}\) sample of plumber's solder, a lead-tin alloy containing \(67 \% \mathrm{Pb}\) by mass and having a density of \(9.4 \mathrm{g} / \mathrm{cm}^{3} ?\)

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