Question

Suppose 0.0934 g of sodium bromide is dissolved in 100 mL of a 11.0 m M aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the sodium bromide is dissolved in it. Be sure your answer has the correct number of significant digits. IM x S ?

Answer 1

To calculate the final molarity of bromide ion, convert the mass of sodium bromide to moles and divide by the volume of the solution. The final molarity is found to be 0.00908 M.

Step-by-step explanation:

The question asks us to calculate the final molarity of bromide anion after dissolving 0.0934 g of sodium bromide in a 100 mL solution containing 11.0 mM silver nitrate. First, we need to convert the mass of sodium bromide to moles using its molar mass. Sodium bromide has a molar mass of 102.89 g/mol, so:

• moles of sodium bromide = 0.0934 g / 102.89 g/mol = 9.08 × 10⁻´ moles

Since the volume of the solution does not change, the total volume remains 100 mL or 0.1 L. To find the molarity of bromide ions, we divide the moles of sodium bromide by the volume of the solution in liters.

• Final molarity of bromide anion = 9.08 × 10⁻´ moles / 0.1 L = 0.00908 M

The final molarity of bromide anions in the solution is 0.00908 M, and this value should be reported with the correct number of significant digits based on the initial measurements provided.