Question

How much potassium bromide, in grams, should be added to water to prepare 0.050 L of a solution with a concentration of 0.125 M?

A. 0.025 g
B. 0.0625 g
C. 0.125 g
D. 0.250 g

Answer 1

To prepare a 0.050 L solution with a concentration of 0.125 M of potassium bromide, you would need to add 0.74375 grams of potassium bromide to water.

Step-by-step explanation:

To calculate the amount of potassium bromide needed for a given volume and concentration of solution, we can use the formula:
moles = volume (L) x concentration (M)

1. Convert the given volume from L to mL: 0.050 L = 50 mL
2. Use the formula to calculate the moles of potassium bromide needed:
   moles = 50 mL x 0.125 M = 6.25 mmol
3. Convert the moles to grams using the molar mass of potassium bromide:
   molar mass of KBr = 39.10 g/mol + 79.90 g/mol = 119.00 g/mol
   grams = 6.25 mmol x 119.00 g/mol = 743.75 mg = 0.74375 g

Therefore, 0.74375 g of potassium bromide should be added to water to prepare 0.050 L of a solution with a concentration of 0.125 M.