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Ammonia reacts with hypobromite, \(\text{OBr}_2\), by the reaction:

\[ \text{2NH}_3 + \text{3OBr}_2 \rightarrow \text{N}_2 + \text{3Br}_2 + \text{3H}_2\text{O} \]

What is the molarity of a hypobromite solution if 1.00 mL of the \(\text{OBr}_2\) solution reacts with 1.69 mg of \(\text{NH}_3\)?

A. 0.01 M
B. 0.02 M
C. 0.03 M
D. 0.04 M

Answer :

Final answer:

To determine the molarity of the hypobromite solution, convert the mass of NH3 to moles, use the stoichiometry to find the moles of OBr2, and divide the moles of OBr2 by the volume of the OBr2 solution in liters to find the molarity.

Explanation:

To determine the molarity of the hypobromite solution, we need to use the stoichiometry of the reaction and the given information. From the balanced equation, we can see that 2 moles of NH3 react with 3 moles of OBr2. Given that 1.00 ml of the OBr2 solution reacts with 1.69 mg of NH3, we can convert the mass of NH3 to moles and then use the stoichiometry to find the moles of OBr2. Finally, we can divide the moles of OBr2 by the volume of the OBr2 solution in liters to find the molarity.


First, let's convert the mass of NH3 to moles.


1.69 mg NH3 * (1 g / 1000 mg) * (1 mol NH3 / 17.03 g) = 9.93 × 10^-5 mol NH3


Now, using the stoichiometry, we can find the moles of OBr2.


(9.93 × 10^-5 mol NH3 * 3 mol OBr2) / 2 mol NH3 = 1.49 × 10^-4 mol OBr2


Finally, we divide the moles of OBr2 by the volume of the OBr2 solution in liters to find the molarity.


Molarity = (1.49 × 10^-4 mol OBr2) / (1.00 ml * 1 L / 1000 ml) = 0.149 M


Therefore, the molarity of the hypobromite solution is 0.149 M.

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