Answer:
a) C Tl = 2 ppb
b) mass Tl = 0.032 μg
c) yes, 0.00164 μg/mL < 0.02 μg/mL, then the water can be safely consumed.
Explanation:
a) C Tl = 0,002 ppm * ( 1000ppb/ppm ) = 2 ppb
b) 16mL * ( L/1000mL ) = 0.016 L
⇒ m Tl = 0,016 L * 0.002 mg/L = 3.2 E-5 mg Tl * ( 1000μg/mg ) = 0.032μg Tl
c) C Tl = 0.00164 μg/mL
∴ according to MCL for 16 mL of sample, the maximum amount allowed is the 0.032μg Tl
⇒ the Tl concentration for 16mL of sample, would be:
⇒ C Tl = 0.032μg Tl / 16 mL = 0.02 μg Tl/mL
∴ the new sample concentration ( 0.0016 μg Tl/mL ) is less than the maximun concentration allowed by MCL ( 0.02μg Tl/mL) for that amount of sample; then the water from that well can be safely consumed.