Question

How many ppm (by wt, mg/kg) W are present in the following?

1) 2.1g of a 0.00312% W ore

2) 28.6g light bulb with a 0.178g filament containing 83.7% tungsten

Answer 1

1).

Given mass of W ore = 2.1 g , percent = 0.00312 %

We have to find concentration in mg / kg .

Calculation of mass of W in 2.1 g of ore in mg.

Mass of W = 2.1 g W ore x 0.00312 g W / 100 g W ore

= 6.55 E-5 g W

= 6.55 E-2 mg W

Conversion of mass of W ore into kg

Mass of W into kg = 2.1 g x 1 kg / 1000 g

= 0.0021 kg

Now concentration of W in W ore in mg/ kg

= 6.55 E-2 mg / 0.0021 kg

= 31.2 mg/kg or

31.2 ppm

2)

Mass of filament in kg = 0.178 x 1 kg / 1000 g

= 0.178 E -3 kg

Mass of W in mg

= 0.178 g x 83.7 g / 100 g

= 0.148986 g

= 148.99 mg

Concentration in ppm

= 148.99 mg / 0.178 E-3 kg

= 837000 ppm