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2 January, 05:49

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.

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  1. 2 January, 09:26
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    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
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