Question

What is the concentration of bromide, in ppm, if 12.41 g MgBr2 is dissolved in 2.55 L water.

Answer 1

Answer:

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

Explanation:

ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.

In our case we have:

mass of MgBr₂ = 12.41 g

volume of water (which is equal to the final solution volume) = 2.55 L

Now we devise the following reasoning:

if         12.41 g of MgBr₂ are dissolved in 2.55 L of water

then         X g of MgBr₂ are dissolved in 1 L of water

X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂

if in         184 g (1 mole) of MgBr₂ we have 160 g of Br⁻

then in   4.867 g of MgBr₂ we have Y g of Br⁻

Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)

4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm