Question

If sea water contains 63 ppm $N{O}_3^{-}$ and its density is 1.01 g/mL, the approximate molarity of sea water is:

• A. $0.001$ M
• B. $0.003$ M
• C. $0.002$ M
• D. $0.004$ M

Answer 1

The correct option is A $0.001$ M

Converting ppm into Molarity

$1ppm=100\times 1000M$

$M=\frac{W}{M.W}\times \frac{1000}{V}=\frac{W}{V}\times \frac{1000}{M.W}$

$=1.1\times \frac{1000}{63}\times 63$

$=\frac{1}{1000}$

$M=0.001M$ .