Question

What is the concentration of bromide, in ppm, if 115.91 g MgBr2 is dissolved in 1.31 L water.

Answer 1

Answer:- Bromide concentration is 80748 ppm.

Solution:- ppm is milligram per liter. Grams of magnesium bromide are given and we are asked to calculate the ppm concentration of bromide ion, So, let's first calculate the grams of bromide and then divide it by the given volume.

Molar mass of Magnesium bromide = $24.305+2(126.90)=278.105$

$115.91gMgBr_2(\frac{1molMgBr_2}{278.105gMgBr_2})(\frac{2molBr}{1molMgBr_2})(\frac{126.90gBr}{1molBr})$

= 105.78 g Br

Let's convert grams of Br to milligrams:

since, 1 g = 1000 mg

$105.78g(\frac{1000mg}{1g})$

= 105780 mg

concentration of Br = $\frac{105780mg}{1.31L}$

= 80748 mg per liter

Since, 1 ppm = 1 mg per liter

So, the bromide concentration is 80748 ppm.