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.
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
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
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