Question

I posted this question before but somebody answered with just "here it is" and no actual answer or breakdown.

The concentration of Br− in a sample of seawater is 6.3 ⋅ 10−4 M. If a liter of seawater has a mass of 1.0 kg, the concentration of Br− is ________ ppm.

Answer 1

Given : molarity of Br- = 6.3 E-4 M

Mass of 1.0 L see water = 1.0 kg

Concentration in ppm = 1mg / 1 kg

Lets find mass of Br- in mg

Calculation of moles of Br- = Molarity of Br- x volume of solution in L

= 6.3 E-4 M x 1.0L

= 6.3 E-4 mol

Mass of Br- in mg = 6.3 E-4 mol x( molar mass of Br-) x (1000 mg / 1 g )

= 6.3 E-4 mol x (79.904 g/mol )x (1000 mg/ 1 g )

= 50.34 mg

Concentration of Br- = 50.34 mg / 1000000 mg

Here we used solution in mg since they have given in kg we converted it to mg

1. kg = 1000000 mg

Lets find concentration of Br- in ppm

Concentration of Br- = 50.34 mg / 1000000 mg

= 5.034 E-5 ppm

Concentration of Br- in ppm = 5.034 E-5