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1.6 g of pyrolusite ore was treated with 50cm^(3) of 1.0 N oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to 250cm^(3) in a flask. 25cm^(3) of this solution when titrated with "0.1 N KMnO"_(4) required 32cm^(3) of the solution. Calculate the percentage of pure MnO_(2) in the ore. |
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Answer» SOLUTION :`25xm^(3)" of OXALIC ACID left required "KMnO_(4)=32cm^(3)" of0.1 M"` `therefore 250 cm^(3)" of the oxalic acid required "KMnO_(4) = 320 cm^(3)" of 0.1 N = 32 cm"^(3) " of 1 N"` `therefore"Oxalic acid used up PYROLUSITE "=(50-32)cm^(3)" of 1 N "=18cm^(3)" of 1 N"` `18cm^(3)" fo 1 N oxalic acid = 18 millieq. of oxalic acid"` It must have reacted with equivalent amount of `MnO_(2)" of pyrolusite"` `"Eq. wt. of "MnO_(2)=(55+32)/(2)=(87)/(2)=43.5` `therefore"18 millieq. Of oxalic acid = 18 millieq. of "MnO_(2)=(18)/(1000)xx43.5g=0.783g` `therefore""%" of pure "MnO_(2)" in pyrolusite "=(0.783)/(1.6)xx100=48.9%` |
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