Question

Ammonia reacts with hypobromite, ${\mathbf{OBr}}^{\mathbf{-}}\mathbf{,}{\mathbf{OBr}}^{\mathbf{-}}$by the reaction ${\mathbf{\text{2NH}}}_{\mathbf{3}}\mathbf{+}\mathbf{30}{\mathbf{Br}}^{\mathbf{-}}\mathbf{\to }{\mathbf{N}}_{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{Br}}^{\mathbf{-}}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$. What is the molarity of a hypobromite solution $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{mL}$of the ${\mathbf{OBr}}^{\mathbf{-}}$solution reacts with$1.69\mathrm{mg}\mathrm{of}{\mathrm{NH}}_3$

Short answer

The molarity of${\mathrm{OBr}}^{-}$is${\mathrm{M}}_{\mathrm{OBr}}=0.15{\mathrm{MOBr}}^{-}$

Step-by-step solution

Define the molarity. 

The total number of moles of solute per litre of solution is known as a solution's molarity. Because, unlike mass, the volume of a system fluctuates with changes in physical conditions, the molality of a solution is affected by changes in physical parameters of the system such as pressure and temperature. Molarity is represented by the letter M, which stands for molar.

The molarity of a solution is defined as one gramme of solute dissolved in one liter of solution.

Calculate the molarity of OBr

$$\begin{gathered} \mathrm{Volume}\mathrm{of}{\mathrm{OBr}}^{-}=1.00\mathrm{mL}=0.001\mathrm{L} \\ \mathrm{mass}\mathrm{of}{\mathrm{NH}}_3=1.69\mathrm{mg}=0.00169\mathrm{g} \\ \mathrm{FM}\mathrm{of}{\mathrm{NH}}_3=117.03\mathrm{g}/\mathrm{mol} \\ \mathrm{Now}\mathrm{use}\mathrm{mole}\mathrm{concept}\mathrm{and}\mathrm{molarity}\mathrm{formula}\mathrm{to}\mathrm{calculate}\mathrm{the}\mathrm{molarity}\mathrm{of}{\mathrm{OBr}}^{-}, \\ {\mathrm{M}}_{{\mathrm{OBr}}^{-}=\frac{0.00169{\mathrm{gNH}}_3\left({\displaystyle \frac{1{\mathrm{molNH}}_3}{17.03{\mathrm{gNH}}_3}}\right)\left({\displaystyle \frac{3{\mathrm{molOBr}}^{-}}{2{\mathrm{molNH}}_3}}\right)}{0.001\mathrm{L}}} \\ =0.15\frac{\mathrm{mol}\mathrm{OBr}}{\mathrm{L}} \\ =0.15{\mathrm{MOBr}}^{-} \\ \mathrm{Thus},\mathrm{the}\mathrm{molarity}\mathrm{of}\mathrm{OBr}\mathrm{is}{\mathrm{M}}_{\mathrm{OBr}}=0.15{\mathrm{MOBr}}^{-} \end{gathered}$$