Question

What is the concentration of bromide, in ppm, if 176.63 g MgBr 2 is dissolved in 7.32 L water. Record your answer in scientific notation to 3 significant figures.

Answer 1

Molar mass of MgBr2 = Molar mass of Mg + 2 * Molar mass of Br = 24.3 + 2 * 79.9 = 184.1 gm/mol

Number of moles of MgBr2 = Mass/Molar mass = 176.63/184.1 = 0.959 moles

Molarity of MgBr2 = Number of moles of MgBr2/Volume of solution (in L) = 0.959/7.32 = 0.131M

One mole of MgBr2 on dissolution will yield 2 moles of Br- and one mole of Mg(2+)

Therefore, concentration of Br- = 2 * 0.131 M = 0.262M

0.262 M/L will be 0.262 * 79.9 g/L = 20.9 g/L

The parts per million answer will be 20.9 * 10^3

Therefore, the final answer in three significant digits and scientific notation will be 2.09 * 10^4 ppm