Question

If ${\text{5}}$ L of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$produce $50$ L of ${{\text{O}}_{\text{2}}}$ at NTP, ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$is:
A. $50$ volume
B. $10$volume
C. $5$volume
D. None of the above

Answer 1

To answer this question we should know what x volume of a substance means. X volume of any substance means that $1$ mL of that substance on decomposition by heating produces x mL of the product. Will convert the all given volumes in mL first. Then according to the definition of x volume will choose the option that produces $50$ L of ${{\text{O}}_{\text{2}}}$.

Complete solution:
First we will convert the given volumes of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ and ${{\text{O}}_{\text{2}}}$ from litre to millilitre as follows:
We know that one liter is equal to a thousand millilitre.
$1$L = $1000$mL
Volume of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$; ${\text{5}}$L = $5000$mL
Volume of${{\text{O}}_{\text{2}}}$; ${\text{50}}$L = $50,000$mL
So, we have to find what volume of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ will give ${\text{50}}$ L ${{\text{O}}_{\text{2}}}$. So, we will compare the volumes given in options with $5000$mL ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$.
So, in option (A), $50$volume of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$means $1$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$will produce $50$mL of ${{\text{O}}_{\text{2}}}$on decomposition by heating. So, $5000$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$will produce,
$1$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$= $50$mL of ${{\text{O}}_{\text{2}}}$
$5000$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ = $250,000$mL of ${{\text{O}}_{\text{2}}}$
So, $50$volume of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ will give $250,000$mL of ${{\text{O}}_{\text{2}}}$ whereas our solution of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ is giving $50,000$mL of ${{\text{O}}_{\text{2}}}$ so, option (A) is not correct.
In option (B), $10$volume of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$means $1$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$will produce $10$mL of ${{\text{O}}_{\text{2}}}$on decomposition by heating. So, $5000$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$will produce,
$1$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$= $10$mL of ${{\text{O}}_{\text{2}}}$
$5000$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ = $50,000$mL of ${{\text{O}}_{\text{2}}}$
So, $10$volume of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ will give $50,000$mL of ${{\text{O}}_{\text{2}}}$ and our solution of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ is also giving $50,000$mL of ${{\text{O}}_{\text{2}}}$ it means our solution of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ is of $10$volume so, option (B) is correct.
So, in option (A), $5$volume of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$means $1$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$will produce $5$mL of ${{\text{O}}_{\text{2}}}$on decomposition by heating. So, $5000$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$will produce,
$1$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$= $5$mL of ${{\text{O}}_{\text{2}}}$
$5000$ mL of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ = $25000$mL of ${{\text{O}}_{\text{2}}}$
So, $5$volume of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ will give $25000$mL of ${{\text{O}}_{\text{2}}}$ whereas our solution of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ is giving $50,000$mL of ${{\text{O}}_{\text{2}}}$ so, option (C) is not correct.
So, if ${\text{5}}$ L of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$produce $50$ L of ${{\text{O}}_{\text{2}}}$ at NTP, ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$is $10$ volume.

Therefore, option (B) $10$ volume is correct.

Note: The NTP is known as normal temperature and pressure. The value of normal temperature in kelvin is$298\,{\text{K}}$. The value of the normal pressure in atm is${\text{1}}\,{\text{atm}}$. One mole of a gas at $298\,{\text{K}}$ temperature and ${\text{1}}\,{\text{atm}}$ pressure occupies $22.4\,{\text{L}}$volume. According to Avogadro number one mole of any substance contains $6.023\, \times \,{10^{23}}$ atoms ions or molecules. NTP condition gives the relation between the number of moles and volume.