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Texts: 22- Since 37 mL of consumption is obtained when a solution containing 250 mg of Na2C2O4 is titrated with a KMnO4 solution with a 1.03 factor, what is the actual concentration of the KMnO4 solution? (Na: 23, C:12, O:16, Mn:55, and K:39 g/mol) a- 0.0222 b- 0.021 c- 0.2 d- 0.0202

Name: texts 22 since 37 ml of consumption is obtained when a solution containing 250 mg of na2c2o4 is titrated with a kmno4 solution with a 103 factor what is the actual concentration of the kmno4 86433
Uploaded: 2022-09-27T06:09:16-08:00
Duration: 1 min 57 s
Channel: Ma Ednelyn Lim
Description: texts 22 since 37 ml of consumption is obtained when a solution containing 250 mg of na2c2o4 is titrated with a kmno4 solution with a 103 factor what is the actual concentration of the kmno4 86433

          Texts: 22- Since 37 mL of consumption is obtained when a solution containing 250 mg of Na2C2O4 is titrated with a KMnO4 solution with a 1.03 factor, what is the actual concentration of the KMnO4 solution? (Na: 23, C:12, O:16, Mn:55, and K:39 g/mol) a- 0.0222 b- 0.021 c- 0.2 d- 0.0202

Added by John D.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Ma Ednelyn Lim

09/27/2022

Step 1

To do this, we need to use the molar mass of Na2C2O4, which is (2*23) + (2*12) + (4*16) = 134 g/mol. The given mass of Na2C2O4 is 250 mg, which is equal to 0.25 g. So, the number of moles of Na2C2O4 is 0.25 g / 134 g/mol = 0.00187 mol. Show more…

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Texts: 22- Since 37 mL of consumption is obtained when a solution containing 250 mg of Na2C2O4 is titrated with a KMnO4 solution with a 1.03 factor, what is the actual concentration of the KMnO4 solution? (Na: 23, C:12, O:16, Mn:55, and K:39 g/mol) a- 0.0222 b- 0.021 c- 0.2 d- 0.0202