What is the concentration of phosphate in the original lake water sample if the

Question

What is the concentration of phosphate in the original lake water sample if the unknown lake water sample in the option had been diluted by taking 25.0 mL of lake water and diluting it to the indicated final volume? The option given is Concentration: 28.7 mg/L Final Vol.: 300.0 mL What is the concentration of phosphate in the original lake water sample if the unknown lake water sample in the option had been diluted by taking 25.0 mL of lake water and diluting it to the indicated final volume? The option given is Concentration: 28.7 mg/L Final Vol.: 300.0 mL The option given is Concentration: 28.7 mg/L Final Vol.: 300.0 mL

Explanation / Answer

I hope I am understanding the question correctly.

Given data is as follows:

Concentration phosphate in diluted sample= 28.7 mg/L

and Final volume= 300 ml= 0.300 L

Volume of Original sample = 300 - 25 = 275 ml = 0.275 L

Now calculate the mass of phophates in 300 ml, final sample:

mass of phosphates = 28.7 mg/L * 0.300 L = 8.61 mg.

This will be same before and after the dilution.

Hence,

Concentration of phosphate in original sample is = mass of phosphate / volume of original sample

Concentration of phosphate in original sample is = 8.61 mg/ 0.275 L = 31.309 mg/L

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