Question

Find the molar concentration (in $mol{L^{ - 1}}$ ) of 96g of ${O_2}$ contained in the vessel of volume 2 liters.
A.16.0
B.1.5
C.4.2
D.24.0

Answer 1

Find the molecular weight of ${O_2}$. Then, find the number of moles of ${O_2}$ by dividing the given weight of ${O_2}$ by its molecular weight. Use the number of moles of ${O_2}$ in finding the molar concentration of 96 g of ${O_2}$.

Formula Used:
To find the molar concentration the formula used is –
Molar concentration $= \dfrac{{{N_m}}}{V}$
where, ${N_m}$ is the number of moles, and
$V$ is the volume

Complete Step by Step Solution:
Molar concentration can also be called molarity. The molar concentration can be defined as the measure of the concentration of the chemical species, in particular of the solute in the terms of amount of the substance per unit volume of solution. The most commonly used unit for molar concentration is moles per liter. It is denoted by $c$ . Mathematically, it is represented by the formula as –
$c = \dfrac{{{N_m}}}{V} \cdots \left( 1 \right)$
where, ${N_m}$ is the number of moles, and
$V$ is the volume
Therefore, according to the question, it is given that, weight of ${O_2}$ is 96 grams and is contained in the volume of vessel 2 liters.
Now, calculating the molecular weight of ${O_2}$
We know that, atomic mass of oxygen is 16. Therefore, the molecular weight of ${O_2}$ is $\Rightarrow 2 \times 16 = 32g/mol$
Now, to find the number of moles we have to divide the weight of ${O_2}$ by its molecular weight. So, -
${N_m} = \dfrac{{96}}{{32}} = 3$
Hence, the number of moles is 3.
From the equation (1), we know that, molar concentration is given by –
$c = \dfrac{{{N_m}}}{V}$
Putting the values of number of moles and volume of vessel which is 2 liters, we get –
$\Rightarrow c = \dfrac{3}{2} \\\ \therefore c = 1.5mol{L^{ - 1}} \\\$
Hence, the molar concentration of 96g of ${O_2}$ contained in the vessel of volume 2 liters is $1.5mol{L^{ - 1}}$

Therefore, the correct option is (B).

Note: The molar concentration depends on the variation of the volume of the solution due mainly to thermal expansion. On small intervals of temperature, the dependence is –
${c_i} = \dfrac{{{c_{{i_1}}}{{\rm T}_0}}}{{1 + \alpha \Delta T}}$
where, ${c_{{i_1}}}{{\rm T}_0}$ is the molar concentration at reference temperature and $\alpha$ is the thermal coefficient of the mixture.