Question

A cold drink bottle contains 200 ml liquid in which $C{O_2}$ is 0.1 molar. Considering $C{O_2}$
as an ideal gas, the volume of the dissolved $C{O_2}$ at STP is:
(A) 22.4 L
(B) 0.224 L
(C) 2.24 L
(D) 0.448 L

Answer 1

To find volume of any gas, we know formula, using mole concept is:
$moles(n) = \dfrac{{{\text{Volume of gas at STP in litres}}}}{{22.4{\text{ }}litres}}$
If we get moles, then we can easily find out the volume of gas. Now, to get moles, we have been given
volume of solution and molarity, so by using molarity formula we can get moles:
$Molarity = \dfrac{{{\text{moles of solute}}}}{{{\text{Volume of solution}}}}$

Complete step by step answer:
We have a cold drink bottle with a volume of 200 ml.
Volume of solution (V) = 200ml = 0.2 Litre
Molarity (M) of solution = 0.1 molar
Now we can find moles easily, using the formula of molarity.
$Molarity = \dfrac{{{\text{moles of solute}}}}{{{\text{Volume of solution}}}}$
$Molarity = \dfrac{{{\text{moles of C}}{{\text{O}}_{\text{2}}}}}{{{\text{Volume of solution}}}}$
Now substitute the values, and then we get:
$0.1 = \dfrac{{{\text{moles of C}}{{\text{O}}_{\text{2}}}}}{{0.2}}$
Taking numbers on one side, and multiplying we can get moles of $C{O_2}$
$\therefore {\text{moles of C}}{{\text{O}}_{\text{2}}} = 0.02$
Now, when we got moles of $C{O_2}$ so we can get volume of gas by using basic formula of calculation
of moles, which can be given below:
$moles(n) = \dfrac{{{\text{Volume of gas at STP in litres}}}}{{22.4{\text{ }}litres}}$
22.4 litres is value at STP, thus substitute 0.02 moles of $C{O_2}$ in above equation, we get:
$0.02 = \dfrac{{{\text{Volume of C}}{{\text{O}}_{\text{2}}}}}{{22.4}}$

On simplifying and calculating, we get:

$${\text{Volume of C}}{{\text{O}}_{\text{2}}} = 0.02 \times 22.4 \\\ \therefore {\text{Volume of C}}{{\text{O}}_{\text{2}}} = 0.448{\text{ litres}} \\\$$

Thus the correct option is D.

Additional information: We know ideal gas equation is:
$PV = nRT$
STP condition means standard temperature and pressure, so the value of Pressure is 1 atm, temperature
is 273K, and R is $0.0821atm.Lmo{l^{ - 1}}{K^{ - 1}}$
Substitute the value into ideal gas equation:
$1 \times V = n \times 0.0821 \times 273$
Rearranging and multiplying values, we can write:
$n = \dfrac{V}{{22.4}}$
This volume (V) is in litres and thus we have seen the derivation of mole formula for ideal gas at STP
conditions.

Note:
Take care of using Volume in litres not in ml, and we should know that Volume of ideal gas at STP.
The condition is 22.4 Litres, from which we can find moles easily. Even while substituting volume value in
molarity formula, it should be in litres, thus we know that molarity is moles of solute present in 1 litre of
Solution.