How many moles of KOH are present in 59.9 mL of 0.711 M KOH? 84.2 mol 42.5 mol 0

Question

How many moles of KOH are present in 59.9 mL of 0.711 M KOH? 84.2 mol 42.5 mol 0.711 mol 4.26 times 10^-2 mol 8.42 times 10^-1 mol The concentration of sulfate in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of the wastewater is added an excess of calcium nitrate, forming the insoluble calcium sulfate (136.1 g/mol) according to the balanced equation given below. The solid calcium sulfate is dried, and its mass is measured to be 0.5413 g. What was the concentration of sulfate in the original wastewater sample? 0.03977 M 0.7361 M 2.514 M 25.14 M 7.367 M

Explanation / Answer

25) no of mole of KOH = M*V = 0.711*59.9 = 42.6 mmole

              = 0.0426 mole

    answer: d

26) no of mole of CaSO4 formed = w/M = 0.5413/136.14 = 0.003977 mole

   no of mole of SO4-2 present in sample = 0.003977 mole

   concentration of SO4-2 = n/V = 0.003977/0.1 = 0.03977 M

answer: a

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