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A sample of drinking water contains 668 ppm of lead. How many grams of lead are there in 100.0 g of this water?

1 Answer

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Answer:

0.0668g

Step-by-step explanation:

Step 1: Given data

Concentration of lead: 668 ppm (mg/kg)

Mass of water: 100.0 g

Step 2: Convert the mass of water to kilograms

We will use the relationship 1 kg = 1,000 g.


100.0g * (1kg)/(1,000g) = 0.1000kg

Step 3: Calculate the mass of lead in 0.1000 kg of water

There are 668 mg of Pb in 1 kg of water.


0.1000kgWater * (668mgPb)/(1kgWater) = 66.8mgPb

Step 4: Convert the mass of Pb to grams

We will use the relationship 1 g = 1,000 mg.


66.8mg * (1g)/(1,000mg) = 0.0668g

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