A solution is prepared by weighing out 0.2351 g of potassium hydrogen phthalate
ID: 551639 • Letter: A
Question
A solution is prepared by weighing out 0.2351 g of potassium hydrogen phthalate (MW = 204.221 g/mol), dissolving it, and diluting to volume with deionized water in a 100.00 mL volumetric flask. A 25.00 mL aliquot is then taken with a volumetric pipet for analysis. Calculate the number of moles of potassium hydrogen phthalate in the aliquot, along with the absolute error in the value. Express your final answer with the correct number of digits.
Is it posible to show the steps as to how to do this question?
Answer given: (2.878±0.005)×10^-4 mol
Explanation / Answer
0.2351 g of potassium hydrogen phthalate in 100 ML
25 mL of aliquot contains potassium hydrogen phthalate = (0.2351/100)x25
= 0.0587 g
Number of moles = weight of potassium hydrogen phthalate /MW of potassium hydrogen phthalate
Moles = 0.0587/204.221
0.0002877
= 2.877x10-4
Absolute error is half of the smallest unit
answer is (2.877±0.005)x10-4
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