Question

If 0.015 mg Pb2+ were added to enough water to make exactly 1 L of solution, the solution would be ____ M in Pb2+

7.2x10-3

1.5x10-5

7.2x10-8

1.5x10-8

7.2x10-5

7.2x103

2. The MCL for Pb2+ is 0.015 ppm this is equivalent to ______

15 mg/L

1.5 mg/L

0.15 mg/L

0.015 mg/L

0.0015 mg/L\]\

Answer 1

Answer – 1) We are given, mass of Pb²⁺ = 0.015 mg, volume = 1.0 L

Molarity of solution = ?

We need to calculate the moles of Pb²⁺

We know,

1 mg = 0.001 g

So,0.015 mg = ?

= 1.5*10^-5 g

moles of Pb²⁺ = 1.5*10^-5 g / 207.2 g.mol^-1

                       = 7.24*10^-8 moles

We know molarity is

Molarity = moles / L

               = 7.24*10^-8 moles / 1.0 L

                = 7.24*10^-8 M

So, the solution would be 7.2*10^-8 M in Pb²⁺

2) [Pb²⁺] = 0.015 ppm

We know

1 ppm = 1 mg/ L

So, 0.015 ppm = 0.015 mg/L

So, the MCL for Pb2+ is 0.015 ppm this is equivalent to 0.015 mg/L