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Chapter 8: Problem 122

When \(500 \mathrm{ml} \mathrm{CO}_{2}\) gas is passed through red hot charcoal, the volume becomes \(700 \mathrm{ml}\). The volume of \(\mathrm{CO}_{2}\) converted into \(\mathrm{CO}\) is (a) \(200 \mathrm{ml}\) (b) \(300 \mathrm{ml}\) (c) \(350 \mathrm{ml}\) (d) \(500 \mathrm{ml}\)

Short Answer

Expert verified
Answer: The volume of \(\mathrm{CO}_{2}\) converted into \(\mathrm{CO}\) is \(300 \mathrm{ml}\).

Step by step solution

01

Write down given quantities

The given quantities in this problem are: - Initial volume of \(\mathrm{CO}_{2} = 500\,\text{ml}\) - Total volume after passing through red hot charcoal \(= 700\,\text{ml}\) We need to find the volume of \(\mathrm{CO}_{2}\) converted into \(\mathrm{CO}\).
02

Find out the volume increase of gas

As we know, the total volume increased from \(500\,\text{ml}\) to \(700\,\text{ml}\), so we will find the difference: $$ \text{Volume increase} = 700\,\text{ml} - 500\,\text{ml} = 200\,\text{ml} $$
03

Analyze the conservation of matter principle

According to the conservation of matter principle, the total number of moles remains constant before and after the reaction. Thus, we have $$ \text{Moles of }\mathrm{CO}_{2}\text{ before} - \text{Moles of }\mathrm{CO}_{2}\text{ after conversion} = \text{Moles of }\mathrm{CO}\text{ formed} $$ As the moles of gas are directly proportional to their volume, we can rewrite the equation in terms of volume: $$ \text{Volume of }\mathrm{CO}_{2}\text{ before} - \text{Volume of }\mathrm{CO}_{2}\text{ after conversion} = \text{Volume of }\mathrm{CO}\text{ after conversion} $$
04

Calculate the volume of \(\mathrm{CO}_{2}\) converted

Using the equation from Step 3: $$ \text{Volume of }\mathrm{CO}_{2}\text{ before} - \text{Volume of }\mathrm{CO}_{2}\text{ after conversion} = 200\,\text{ml} $$ Substituting the initial volume of \(\mathrm{CO}_{2}\), we get: $$ 500\,\text{ml} - \text{Volume of }\mathrm{CO}_{2}\text{ after conversion} = 200\,\text{ml} $$ Solving for the volume of \(\mathrm{CO}_{2}\) after conversion: $$ \text{Volume of }\mathrm{CO}_{2}\text{ after conversion} = 500\,\text{ml} - 200\,\text{ml} = 300\,\text{ml} $$ So, the volume of \(\mathrm{CO}_{2}\) converted into \(\mathrm{CO}\) is \(300\,\text{ml}\). The correct option is (b) \(300 \mathrm{ml}\).

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Most popular questions from this chapter

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) \(2 / 5\)

A volume of \(10 \mathrm{~m}\) l chlorine gas combines with \(25 \mathrm{~m}\) l of oxygen gas to form \(10 \mathrm{ml}\) of a gaseous compound. If all the volumes are measured at the same pressure and temperature, what is the molecular formula of compound formed? (a) \(\mathrm{Cl}_{2} \mathrm{O}\) (b) \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) (c) \(\mathrm{ClO}_{2}\) (d) \(\mathrm{Cl}_{2} \mathrm{O}_{5}\)

A volume of \(200 \mathrm{ml}\) of oxygen is added to \(100 \mathrm{ml}\) of a mixture containing \(\mathrm{CS}_{2}\) vapour and \(\mathrm{CO}\), and the total mixture is burnt. After combustion, the volume of the entire mixture is \(245 \mathrm{ml}\). Calculate the volume of the oxygen that remains (a) \(67.5 \mathrm{ml}\) (b) \(125.0 \mathrm{ml}\) (c) \(200.0 \mathrm{ml}\) (d) \(100.0 \mathrm{ml}\)

How many grams of \(90 \%\) pure \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) can be produced from \(250 \mathrm{~g}\) of \(95 \%\) pure \(\mathrm{NaCl} ?\) (a) \(640.6 \mathrm{~g}\) (b) \(288.2 \mathrm{~g}\) (c) \(259.4 \mathrm{~g}\) (d) \(320.3 \mathrm{~g}\)

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) \(\mathrm{X}_{4} \mathrm{Y}\)
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