Question

The molarity of solution and weight of sodium bromate necessary to prepare 85.5 mL of 0.672N solution when the half-cell reaction is $\text{Br}{{\text{O}}_{3}}^{-}+\text{6}{{\text{H}}^{+}}+\text{6}{{\text{e}}^{-}}\to \text{B}{{\text{r}}^{-}}+\text{3}{{\text{H}}_{2}}\text{O}$, are respectively:
A. M= 0.112 M, Weight= 1.446 mg
B. M= 0.224 M, Weight= 2.892 mg
C. M= 0.112 M, Weight= 1.446 mg
D. M= 1.112 M, Weight= 4.338 mg

Answer 1

Hint: The question is based on volumetric methods. The dealing in volume, moles and equivalents is done in volumetric section. The relation between molarity and normality is Normality= Molarity$\times $n-factor. We need to find the n-factor of $\text{Br}{{\text{O}}_{3}}^{-}$ and by its moles and equivalents. The formula of molarity is $\dfrac{\text{number of moles}}{\text{volume (in mL)}}\times 1000$. The question will be done thereafter.

Complete step by step answer:
Step (1)- Find the n-factor of sodium bromate or $\text{Br}{{\text{O}}_{3}}^{-}$, to find the n-factor we are given the half-cell reaction; $\text{Br}{{\text{O}}_{3}}^{-}+\text{6}{{\text{H}}^{+}}+\text{6}{{\text{e}}^{-}}\to \text{B}{{\text{r}}^{-}}+\text{3}{{\text{H}}_{2}}\text{O}$, the number of electrons involved in the reaction gives us the account of n-factor of $\text{Br}{{\text{O}}_{3}}^{-}$. So, the electrons involved in this reaction are 6 electrons. Thus, the n-factor of $\text{Br}{{\text{O}}_{3}}^{-}$ is 6.

Step (2)- Connect molarity with normality; in molarity we deal with moles and in normality we deal with equivalents. Normality is Molarity multiplied with n-factor or Normality= Molarity$\times $n-factor. We have normality and n-factor, so, molarity can be directly found by just dividing the terms. The formulation will be $\dfrac{0.672}{6}$ which is equal to 0.112 M.

Step (3)- Find the weight of $\text{Br}{{\text{O}}_{3}}^{-}$, the moles of $\text{Br}{{\text{O}}_{3}}^{-}$need to be obtained by the formula of molarity. Molarity is the number of moles of a substance in one litre of volume. It can be mathematically expressed as; Molarity= $\dfrac{\text{number of moles}}{\text{volume (in mL)}}\times 1000$
In the question, the molarity is 0.112 M (calculated above) and volume of the solution as 85.5$\text{mL}$. Put the values in the formula of molarity and obtain the number of moles of $\text{Br}{{\text{O}}_{3}}^{-}$. The expression will be like $0.112=\dfrac{\text{number of moles}}{85.5}\times 1000$, the number of moles of $\text{Br}{{\text{O}}_{3}}^{-}$ are obtained as 0.0095 moles.

Step (4)- Find the molecular mass- The molecular weight of sodium bromate ($\text{NaBr}{{\text{O}}_{3}}$) is
- Atomic mass of sodium is 23 grams.
- Atomic mass of oxygen is 16 grams.
- Atomic mass of bromine is 80 grams.
The molecular weight of sodium bromate ($\text{NaBr}{{\text{O}}_{3}}$) will be 23+$3\times 16$+80 is equal to 151 grams.

Step (5)- We know that moles are equal to weight of substance by molecular weight or moles=$\dfrac{\text{weight of substance}}{\text{molecular mass}}$, so the weight of ($\text{NaBr}{{\text{O}}_{3}}$) will be moles$\times $molecular mass which is equal to 1.446 grams.
-The weight of sodium bromate ($\text{NaBr}{{\text{O}}_{3}}$) is 1.446 grams and Molarity of the solution is 0.112 M.
Hence, the correct answer is option A.

Note:
The answer to the question can be obtained directly by one formula;
Normality=$\dfrac{\dfrac{\text{weight of substance}}{\text{molar mass}}}{\dfrac{\text{volume}}{\text{n-factor}}}\times 1000$; where n-factor is 6, molar mass is 151 grams, volume is 85.5 ml and normality is 0.672N. The weight of sodium bromate will be $0.672=\dfrac{\dfrac{\text{weight of substance}}{151}}{\dfrac{85.5}{6}}\times 1000$; weight of sodium bromate is 1.446 grams.