Question

A 20.0 mL solution of $N{{a}_{2}}S{{O}_{3}}$ required 30 mL of 0.01M ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ solution for the oxidation of $N{{a}_{2}}S{{O}_{4}}$. Hence, molarity of $N{{a}_{2}}S{{O}_{3}}$ solution is:
A. 0.015M
B. 0.045M
C. 0.030M
D. 0.0225M

Answer 1

This question belongs to the stoichiometry. Stoichiometry is the calculation of reactants and products in a chemical reaction. This is based on the Law of Conservation of Mass where the total mass of reactants is equal to the total mass of the products. It can be effectively used to calculate one of the reactants or products with the comparison of other side compounds.

Complete step by step solution:
First of all, a reaction is to be written to compare between the reactants and products
${{K}_{2}}C{{r}_{2}}{{O}_{7}}+4{{H}_{2}}S{{O}_{4}}+3N{{a}_{2}}S{{O}_{3}}\to {{K}_{2}}S{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+3N{{a}_{2}}S{{O}_{4}}+4{{H}_{2}}O$

Now, with the help of equation,
1 mole of Potassium dichromate corresponds to 3 moles of Sodium sulphite.
Hence, millimoles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ = 3 × 0.01 = 0.3 mole
Number of millimoles of $N{{a}_{2}}S{{O}_{3}}$= 3 × 0.3= 0.9 mole
Therefore, molarity of $N{{a}_{2}}S{{O}_{3}}$ = $\dfrac{0.9}{20.0}$ = 0.045 M
So, the correct answer is “Option B”.

Note: This question can be done by a different formula
${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$

Where, the N= M × n-factor. Just remember to find the millimoles of the given compound if you do not want to write the equation. Otherwise, this formula cannot be directly applied, if you will apply it directly then, the answer will change to 0.015M. You may ask that we have to find molarity not normality but the millimoles can only be found with the help of n-factor.