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Chapter 8: Problem 21

Twenty molecules of \(\mathrm{SO}_{3}\) will weigh as much as \(\ldots . .\) molecules of oxygen. (a) 100 (b) 50 (c) 15 (d) 8

Short Answer

Expert verified
Answer: 50 oxygen molecules

Step by step solution

01

Find the molecular weight of \(\mathrm{SO}_{3}\)

First, we need to find the molecular weight of one \(\mathrm{SO}_{3}\) molecule. We know that the atomic weight of sulfur (S) is 32 u (atomic mass units) and that of oxygen (O) is 16 u. Since there are three oxygen atoms in \(\mathrm{SO}_{3}\), the molecular weight of \(\mathrm{SO}_{3}\) will be \(32 + 3 \times 16 = 32 + 48 = 80\) u.
02

Calculate the weight of 20 molecules of \(\mathrm{SO}_{3}\)

Now that we know the molecular weight of one \(\mathrm{SO}_{3}\) molecule, we can find the weight of 20 molecules: \(20 \times 80 = 1600\) u.
03

Find the molecular weight of oxygen

Oxygen gas exists as a diatomic molecule, meaning that there are two oxygen atoms in one molecule. As the atomic weight of oxygen is 16 u, the molecular weight of oxygen gas (\(\mathrm{O}_{2}\)) will be \(2 \times 16 = 32\) u.
04

Determine the number of oxygen molecules

We have the weight of 20 \(\mathrm{SO}_{3}\) molecules as 1600 u. We want to find out how many molecules of oxygen will have the same weight. Let's call this number of oxygen molecules x. So, the equation we need to solve is: \(x \times 32 = 1600\). To find the value of x, divide both sides of the equation by 32: \(x = \frac{1600}{32} = 50\). Therefore, the answer is (b) 50 oxygen molecules will have the same weight as twenty \(\mathrm{SO}_{3}\) molecules.

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Most popular questions from this chapter

Upon heating a litre of semi-molar \(\mathrm{HCl}\) solution, \(2.675 \mathrm{~g}\) of hydrogen chloride is lost and the volume of the solution shrinks to \(750 \mathrm{ml}\). The molarity of resultant solution is (a) \(0.569 \mathrm{M}\) (b) \(0.5 \mathrm{M}\) (c) \(0.42 \mathrm{M}\) (d) \(1.707 \mathrm{M}\)

A gaseous alkane is exploded with oxygen. The volume of \(\mathrm{O}_{2}\) for complete combustion to the volume of \(\mathrm{CO}_{2}\) formed is in \(7: 4\) ratio. The molecular formula of alkane is (a) \(\mathrm{CH}_{4}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{8}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{10}\)

A compound contains elements \(X\) and \(\mathrm{Y}\) in \(1: 4\) mass ratio. If the atomic masses of \(X\) and \(Y\) are in \(1: 2\) ratio, the empirical formula of compound should be (a) \(\mathrm{XY}_{2}\) (b) \(\mathrm{X}_{2} \mathrm{Y}\) (c) \(\mathrm{XY}_{4}\) (d) \(\mathrm{X}_{4} \mathrm{Y}\)

A mixture of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) is caused to react in a closed container to form \(\mathrm{NH}_{3} .\) The reaction ceases before either reactant has been totally consumed. At this stage, \(2.0 \mathrm{moles}\) each of \(\mathrm{N}_{2}, \mathrm{H}_{2}\) and \(\mathrm{NH}_{3}\) are present. The moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) present originally were, respectively, (a) 4 and 4 moles (b) 3 and 5 moles (c) 3 and 4 moles (d) 4 and 5 moles

When a certain amount of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), is burnt completely, \(7.04 \mathrm{~g} \mathrm{CO}_{2}\) is formed. What is the mass of \(\mathrm{H}_{2} \mathrm{O}\) formed, simultaneously? (a) \(1.62 \mathrm{~g}\) (c) \(6.48 \mathrm{~g}\) (c) \(3.24 \mathrm{~g}\) (d) \(2.28 \mathrm{~g}\)
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