Question

Volume occupied by $4.4g$ of $C{O_2}$ is:
A. $2240$ lit at STP
B. $2.24$ lit at STP
C. $22.4$ lit at STP
D. $224$ lit at STP

Answer 1

Hint:The relation between the number of moles, the weight of the substance given and the volume of the substance provided plays a key factor in solving various problems regarding the stoichiometry and has proven helpful in the chemical laboratories for the chemical analysis and physical calculations.

Complete step by step answer:
Let the number of moles of carbon dioxide be $x$ .
As per the question, the given weight of carbon dioxide is = $w = 4.4g$
The molecular weight of the carbon dioxide is needed to be calculated. The molecular weight of any substance is the individual sum of the constituent atoms that form the compound. In case of carbon dioxide, the atomic masses of the individual atoms are as follows:
$C = 12g$
$O = 16g$
Thus, molecular weight of $C{O_2} = {M_w} = 12 + (2 \times 16) = 44g$
Let us determine the number of moles of carbon dioxide present in $4.4g$ of $C{O_2}$ .
$x = \dfrac{w}{{{M_w}}} = \dfrac{{4.4}}{{44}} = 0.1moles$
Now, we also know that the relation between the number of moles and the volume of the given substance is as follows:
$x = \dfrac{V}{{22.4(l)}}$
Where, $x = $ number of moles
$V = $ Volume of carbon dioxide = ?
The volume occupied by one mole of any gas at standard conditions is equal to $22.4l$ .
Now, substituting the values and solving, we have:
$0.1 = \dfrac{V}{{22.4(l)}} \Rightarrow V = 2.24l$

Thus, the correct option is B. $2.24$ lit at STP.

Note: The relation between the number of moles, weight of the given substance, volume of the given amount of substance and the number of particles present in the given amount of substance can be shown as follows:
$n = \dfrac{w}{{{M_w}}} = \dfrac{V}{{22.4(l)}} = \dfrac{N}{{{N_A}}}$
Where, $n = $ number of moles
$w = $ Given weight
${M_w} = $ Atomic/ molecular weight
$V = $ Given volume
$N = $ Number of particles (atoms, ions, molecules, etc)
${N_A} = $ Avogadro number