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(i) What is the mass of sodium bromate and molarity of the solution necessary to prepare 85.4 mL of 0.072 N solution when the half-Sreaction is $\text{Br}{{\text{O}}^{-}}\text{ + 6}{{\text{H}}^{+}}\text{ + 6}{{\text{e}}^{-}}\text{ }\to \text{ B}{{\text{r}}^{-}}\text{ + 3}{{\text{H}}_{2}}\text{O?}$(ii) What would be the mass as well as molarity if the half-cell reaction is $\text{2BrO}_{3}^{-}\text{ + 12}{{\text{H}}^{+}}\text{ + 10}{{\text{e}}^{-}}\text{ }\to \text{ B}{{\text{r}}_{2}}\text{ + 6}{{\text{H}}_{2}}\text{O}$?

Last updated date: 19th Sep 2024
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Hint: Normality is the ratio of the equivalent weight of the solute to the volume in litre. Whereas to find molarity can be calculated by the relationship of normality and molarity i.e. normality = product of n and molarity. Here, n is the total loss or gain of the electron.

(i)- In the given question, we have to find the mass of the sodium bromate and the molarity of the solution with the given reaction.
- As we know that the molecular formula of sodium bromate is \begin{align} & \text{NaBr}{{\text{O}}_{3}} \\ & \\ \end{align} so the molecular mass will be $23\text{ + 79}\text{.9 + 3 }\times \text{ 16 }\sim \text{ 151g}$.
- Now, as we know that the relationship between normality and molarity is
$\text{Normality = n }\times \text{ Molarity}$
- So, here the value of n will be 6 because in the given reaction there are 6 electrons used.
$\text{Br}{{\text{O}}^{-}}\text{ + 6}{{\text{H}}^{+}}\text{ + 6}{{\text{e}}^{-}}\text{ }\to \text{ B}{{\text{r}}^{-}}\text{ + 3}{{\text{H}}_{2}}\text{O}$
- Now, by applying the above formula we will get the value of molarity i.e. $\dfrac{\text{Normality}}{\text{n}}\text{ }=\ \dfrac{0.672}{6}\text{ = 0}\text{.112M}$
- Now, the mass of sodium bromate will be calculated by the formula of molarity i.e.
$\text{Molarity = }\dfrac{\text{Mass }\times \text{ 1000}}{\text{Molar mass }\times \text{ volume in mL}}$
- Or it can also be written as
$\text{Mass = Molarity }\times \text{ molar mass }\times \text{ volume in ml }\times \text{ 1}{{\text{0}}^{3-}}$
$\text{Mass = 0}\text{.112 }\times \text{ 151 }\times \text{ 85}\text{.5 }\times \text{ 1}{{\text{0}}^{3-}}\text{ = 1}\text{.446g}$.
Therefore, molarity is 0.112M and mass of sodium bromate is 1.446g.

(ii)- Now, in this reaction 2 moles of sodium bromate is used which consume 10 electrons. So, one mole will consume 5 electrons.
$\text{2BrO}_{3}^{-}\text{ + 12}{{\text{H}}^{+}}\text{ + 10}{{\text{e}}^{-}}\text{ }\to \text{ B}{{\text{r}}_{2}}\text{ + 6}{{\text{H}}_{2}}\text{O}$
- To calculate the molarity of the solution we will apply the relationship between molarity and normality i.e.
$\text{Normality = n }\times \text{ Molarity}$
- As it is given that the value of normality is 0.672N and the value of n is 5 because 5 electrons participate in the reaction.
- So, the molarity is equal to:
$\dfrac{\text{Normality}}{\text{n}}\text{ }=\ \dfrac{0.672}{5}\text{ = 0}\text{.1344M}$
- Now, the mass of sodium bromate will be calculated by the formula of molarity i.e.
$\text{Molarity = }\dfrac{\text{Mass }\times \text{ 1000}}{\text{Molar mass }\times \text{ volume in mL}}$
- Or it can also be written as:
$\text{Mass = Molarity }\times \text{ molar mass }\times \text{ volume in ml }\times \text{ 1}{{\text{0}}^{3-}}$
$\text{Mass = 0}\text{.1344 }\times \text{ 151 }\times \text{ 85}\text{.5 }\times \text{ 1}{{\text{0}}^{3-}}\text{ = 1}\text{.735g}$.
Therefore, molarity is 0.1344M and mass of sodium bromate is 1.735g.

Note: The molarity and molality are different from each other. Molarity is dependent on the temperature and changes with change in the temperature whereas molality is independent of the temperature.