Question

Phosphorus is present in seawater to the extent of 0.07 ppm by mass. You may want to reference (Page) Section 18.3 while completing this problem. Part A Assuming that the phosphorus is present as dihydrogenphosphate, H2PO43−, calculate the corresponding molar concentration of phosphate in seawater. The density of seawater is 1.025 g/mL.

Answer 1

Answer:

7.22x10⁻⁷ M

Explanation:

The unity ppm means parts per million, thus, the concentration of phosphorus is 0.07 g per 1 million g of seawater. Because the density of seawater is 1.025 g/mL, the concentration of phosphorus is:

0.07g/1million g * 1.025 g/mL = 0.07175 g/ 1 million mL

1 million mL = 1,000 L, thus:

Concentration of phosphorus = 0.07g/1000 L = 7.0x10⁻⁵ g/L

The molar mass of dyhydrogenphosphate is:

2*1 g/mol of H + 1*31 g/mol of P + 4*16 g/mol of O = 97 g/mol

The molar concentration is the mass concentration divided by the molar mass:

7.0x10⁻⁵/97 = 7.22x10⁻⁷ M