Question

Chlorine dioxide $Cl{O_2}$ is used nowadays for water treatment rather than $C{l_2}$.$Cl{O_2}$ is obtained by passing $C{l_2}$(g) into a concentrated solution of sodium chlorite ($NaCl{O_2}$). The reaction gives 90% yield. If x moles of $Cl{O_2}$ is produced in 3.48 L of 2.0 M $NaCl{O_2}$(aq), then calculate the value of 10x.
A. 68
B. 60
C. 74
D. None of these

Answer 1

Hint: We have to determine the moles of $Cl{O_2}$. First we must know about molarity. Therefore, molarity is the number of moles of a solute per liter of a solution. So, we can determine the total number of moles by multiplying molarity of a solute to the total volume of the solution in liters.

Complete step by step answer:
To solve the above equation it is important for us to derive the equation from the above question:
$NaCl{O_2} + C{l_2} \to NaCl + Cl{O_2}$
The main equation:
Now, let’s balance the above equation:
$2NaCl{O_2} + C{l_2} \to 2NaCl + 2Cl{O_2}$
Then, let’s derive the number of moles of $Cl{O_2}$ = the number of moles of $Cl{O_2}$
Now, let’s substitute the values of the following in their respective places.
Formula used:
\[Moles{\text{ }}of{\text{ }}NaCl{O_2} = {\text{ }}Molarity\;Volume{\text{ }}in{\text{ }}Litres\]
$M \times {V_L} = 2 \times 3.78$
$M \times {V_L} = 7.56$
This, 7.56 is the formation of $NaCl{O_2}$
As it is mentioned to us in the question that the actual yield of $Cl{O_2}$ is 90% then let’s again determine the value of moles of $Cl{O_2}$
So, moles of $Cl{O_2}$ with 90% yield =$\dfrac{{7.56 \times 90}}{{100}} = 6.8$
To determine the value of ‘x’ we have to find the value of the respective moles of $Cl{O_2}$.
Don’t forget that the value which we derived earlier 6.8 is equal to x mol of $Cl{O_2}$
${\text{6}}{\text{.8 = x mol of Cl}}{{\text{O}}_{\text{2}}}$
Now, to derive the value of 10x mol of $Cl{O_2}$ $ = 10 \times 6.8 = 6.8$

Note:
We must remember that the number of moles is equal to the product of molarity of a solute and volume of a solution measured in liters. And do not confuse the two words molarity and molality. Therefore, Molarity is the ratio of the moles of a solute to the total liters of a solution. Molality, on the other hand, is the ratio of the moles of a solute to the kilograms of a solvent.