
The ratio of molar volume to atomic volume for 1 mole of hydrogen is (Take size of hydrogen molecule to be 0.5$\mathop {\text{A}}\limits^0 $):
$
A. {\text{ 7}}{\text{.1}} \times {10^4} \\
B. {\text{ 7}}{\text{.1}} \times {10^6} \\
C. {\text{ 7}}{\text{.1}} \times {10^{10}} \\
D. {\text{ 7}}{\text{.1}} \times {10^8} \\
$
Answer
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Hint: The molar volume of a gas is the volume occupied by the 1 mole of that gas at standard temperature and pressure while the atomic volume is the volume occupied by the same 1 mole of gas at room temperature.
Complete step-by-step answer:
We are given a hydrogen molecule and we need to find the ratio of its molar volume and the atomic volume.
These two volumes have different meanings. Molar volume is defined as the volume occupied by 1 mole of the molecules of a gas at STP i.e. standard temperature and pressure while the atomic volume is the volume occupied by 1 mole of a gas at room temperature.
Every 1 mole of a gas occupies the same volume at STP which is 22.4 litres. So, we have the value of molar volume for hydrogen $ = 22.4l = 22.4 \times {10^{ - 3}}{m^3}$.
Now atomic volume can be calculated from the radius of the molecule in the following way. The radius of the hydrogen molecule is given 0.5$\mathop {\text{A}}\limits^0 $.
Therefore atomic volume of hydrogen $ = \dfrac{4}{3}\pi {r^3}{N_A}$
Now $r = 1\mathop {\text{A}}\limits^0 = {10^{ - 10}}m$ and ${N_A}$ is the Avogadro’s number (${N_A} = 6.022 \times {10^{23}}$). Using these values, we can get the atomic volume.
Atomic volume of hydrogen
$
= \dfrac{4}{3} \times 3.14 \times {\left( {0.5 \times {{10}^{ - 10}}} \right)^3} \times 6.022 \times {10^{23}} \\
= 3.15 \times {10^{ - 7}}{m^3} \\
$
Now we can calculate the ratio of the molar volume and atomic volume as follows:
$
\dfrac{{{\text{Molar volume}}}}{{{\text{Atomic volume}}}} = \dfrac{{22.4 \times {{10}^{ - 3}}{m^3}}}{{3.15 \times {{10}^{ - 7}}{m^3}}} \\
= 7.1 \times {10^4} \\
$
Hence, the correct answer is option A.
Note: We notice that the ratio of molar volume to atomic volume is large for hydrogen molecules. The consequence of this is that molecules are separated from each other at large distances as compared to the size of the hydrogen atom.
Complete step-by-step answer:
We are given a hydrogen molecule and we need to find the ratio of its molar volume and the atomic volume.
These two volumes have different meanings. Molar volume is defined as the volume occupied by 1 mole of the molecules of a gas at STP i.e. standard temperature and pressure while the atomic volume is the volume occupied by 1 mole of a gas at room temperature.
Every 1 mole of a gas occupies the same volume at STP which is 22.4 litres. So, we have the value of molar volume for hydrogen $ = 22.4l = 22.4 \times {10^{ - 3}}{m^3}$.
Now atomic volume can be calculated from the radius of the molecule in the following way. The radius of the hydrogen molecule is given 0.5$\mathop {\text{A}}\limits^0 $.
Therefore atomic volume of hydrogen $ = \dfrac{4}{3}\pi {r^3}{N_A}$
Now $r = 1\mathop {\text{A}}\limits^0 = {10^{ - 10}}m$ and ${N_A}$ is the Avogadro’s number (${N_A} = 6.022 \times {10^{23}}$). Using these values, we can get the atomic volume.
Atomic volume of hydrogen
$
= \dfrac{4}{3} \times 3.14 \times {\left( {0.5 \times {{10}^{ - 10}}} \right)^3} \times 6.022 \times {10^{23}} \\
= 3.15 \times {10^{ - 7}}{m^3} \\
$
Now we can calculate the ratio of the molar volume and atomic volume as follows:
$
\dfrac{{{\text{Molar volume}}}}{{{\text{Atomic volume}}}} = \dfrac{{22.4 \times {{10}^{ - 3}}{m^3}}}{{3.15 \times {{10}^{ - 7}}{m^3}}} \\
= 7.1 \times {10^4} \\
$
Hence, the correct answer is option A.
Note: We notice that the ratio of molar volume to atomic volume is large for hydrogen molecules. The consequence of this is that molecules are separated from each other at large distances as compared to the size of the hydrogen atom.
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