
When equal masses of methane and sulphur dioxide are taken, then the ratio of their molecule is:
A. 1:1
B. 1:2
C. 2:1
D. 4:1
Answer
561.9k+ views
Hint: The number of moles is obtained by dividing mass of the substance by molar or molecular mass and the number of molecules is calculated by multiplying moles and Avogadro’s number. Then we’ll calculate the ratio of their molecules.
Complete Step by step answer: A mole is defined as the amount of a substance which constitutes different elementary particles like atoms, molecules or ions equal to Avogadro’s number i.e., $6.022 \times {10^{22}}$ . Avogadro’s number is defined as the number of atoms present in 12 g of C-12 isotope. A molar mass of a chemical compound is defined as the mass of a part of that compound divided by the amount of substance or number of moles in it. Let the molar mass of methane be ${m_1}$, molar mass of sulphur dioxide be ${m_2}$, number of moles of methane be ${n_1}$ and number of moles of sulphur dioxide be ${n_2}$.
Now, we’ll calculate molar mass of methane $\left( {C{H_4}} \right)$ i.e.,${m_1} = 1 \times C + 4 \times H$
${m_1} = 1 \times 12 + 4 \times 1$ as molar or molecular mass of carbon is 12g and hydrogen is 1g.
$\Rightarrow {m_1} = 12 + 4$
$\Rightarrow {m_1} = 16g$
Molar mass of sulphur dioxide $\left( {S{O_2}} \right)i.e.,{m_2} = 1 \times S + 2 \times O$
${m_2} = 1 \times 32 + 2 \times 16$ as molar mass of sulphur is 32g and oxygen is 16g.
$\Rightarrow {m_2} = 32 + 32$
$\Rightarrow {m_2} = 64g$
${n_1} = \dfrac{{mass{\text{ }}of{\text{ }}methane}}{{{m_1}}}$
$\Rightarrow {n_1} = \dfrac{m}{{16}}$ [eqn.1]
${n_2} = \dfrac{{mass{\text{ }}of{\text{ }}sulphur{\text{ }}dioxide}}{{{m_2}}}$
$\Rightarrow {n_2} = \dfrac{m}{{64}}$ since equal masses are taken. [eqn.2]
We divide eqn.1 by 2,
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\dfrac{m}{{16}}}}{{\dfrac{m}{{64}}}}$
$\Rightarrow \dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{64}}{{16}}$
$\Rightarrow \dfrac{{{n_1}}}{{{n_2}}} = \dfrac{4}{1}$
$\Rightarrow \dfrac{{Number{\text{ }}of{\text{ }}molecules{\text{ }}of{\text{ }}methane}}{{Number{\text{ }}of{\text{ }}molecules{\text{ }}of{\text{ }}sulphur{\text{ }}dioxide}}$ =$\dfrac{{{n_1} \times 6.022 \times {{10}^{22}}}}{{{n_2} \times 6.022 \times {{10}^{22}}}}$
i.e. $\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{4}{1}or4:1$
Therefore, option D is correct.
Note: We should remember that the number of molecules is obtained by multiplying the number of moles and Avogadro’s number. The number of moles is inversely proportional to molar mass of the compound. Thus, the another way to find the ratio of their molecule is on dividing the molar mass of sulphur dioxide by methane
Complete Step by step answer: A mole is defined as the amount of a substance which constitutes different elementary particles like atoms, molecules or ions equal to Avogadro’s number i.e., $6.022 \times {10^{22}}$ . Avogadro’s number is defined as the number of atoms present in 12 g of C-12 isotope. A molar mass of a chemical compound is defined as the mass of a part of that compound divided by the amount of substance or number of moles in it. Let the molar mass of methane be ${m_1}$, molar mass of sulphur dioxide be ${m_2}$, number of moles of methane be ${n_1}$ and number of moles of sulphur dioxide be ${n_2}$.
Now, we’ll calculate molar mass of methane $\left( {C{H_4}} \right)$ i.e.,${m_1} = 1 \times C + 4 \times H$
${m_1} = 1 \times 12 + 4 \times 1$ as molar or molecular mass of carbon is 12g and hydrogen is 1g.
$\Rightarrow {m_1} = 12 + 4$
$\Rightarrow {m_1} = 16g$
Molar mass of sulphur dioxide $\left( {S{O_2}} \right)i.e.,{m_2} = 1 \times S + 2 \times O$
${m_2} = 1 \times 32 + 2 \times 16$ as molar mass of sulphur is 32g and oxygen is 16g.
$\Rightarrow {m_2} = 32 + 32$
$\Rightarrow {m_2} = 64g$
${n_1} = \dfrac{{mass{\text{ }}of{\text{ }}methane}}{{{m_1}}}$
$\Rightarrow {n_1} = \dfrac{m}{{16}}$ [eqn.1]
${n_2} = \dfrac{{mass{\text{ }}of{\text{ }}sulphur{\text{ }}dioxide}}{{{m_2}}}$
$\Rightarrow {n_2} = \dfrac{m}{{64}}$ since equal masses are taken. [eqn.2]
We divide eqn.1 by 2,
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\dfrac{m}{{16}}}}{{\dfrac{m}{{64}}}}$
$\Rightarrow \dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{64}}{{16}}$
$\Rightarrow \dfrac{{{n_1}}}{{{n_2}}} = \dfrac{4}{1}$
$\Rightarrow \dfrac{{Number{\text{ }}of{\text{ }}molecules{\text{ }}of{\text{ }}methane}}{{Number{\text{ }}of{\text{ }}molecules{\text{ }}of{\text{ }}sulphur{\text{ }}dioxide}}$ =$\dfrac{{{n_1} \times 6.022 \times {{10}^{22}}}}{{{n_2} \times 6.022 \times {{10}^{22}}}}$
i.e. $\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{4}{1}or4:1$
Therefore, option D is correct.
Note: We should remember that the number of molecules is obtained by multiplying the number of moles and Avogadro’s number. The number of moles is inversely proportional to molar mass of the compound. Thus, the another way to find the ratio of their molecule is on dividing the molar mass of sulphur dioxide by methane
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

