1.

30 mL of a H_(2)O_(2) solution after acidification required 30 mL of N/10 KMnO_(4)solution for complete oxidation . Calculate the percentage and volume strength of H_(2)O_(2) solution.

Answer» <html><body><p></p>Solution :To determine the normality of `H_(2)O_(2)` solution . From the given <a href="https://interviewquestions.tuteehub.com/tag/data-25577" style="font-weight:bold;" target="_blank" title="Click to know more about DATA">DATA</a>,<br/>for `H_(2)O_(2), V_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)=30 mL, N_(1)=?` <br/>For `KMnO_(4), V_(2)=30 mL, N_(2)=N//10` <br/>Applying normallity equation , `N_(1)V_(1)=N_(2)V_(2) , i.e., 30xxN_(1)=30xx1//10 therefore N_(1)=0.1 N` <br/>Thus, the normality of `H_(2)O_(2)` solution =0.1 N <br/>Step 2. To determine the percentage <a href="https://interviewquestions.tuteehub.com/tag/strength-1229153" style="font-weight:bold;" target="_blank" title="Click to know more about STRENGTH">STRENGTH</a> of `H_(2)O_(2)` solution, <br/>We know that, `H_(2)O_(2) to 2H^(+) + O_(2) + 2e^(-)""therefore"` Eq. wt. of `H_(2)O_(2) =34//2=17` <br/>Hence, strength of `H_(2)O_(2)=(1.7xx100)/(1000)=0.17%` <br/>Step 3. To determinethe <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> strength of `H_(2)O_(2)` solution. <br/>Consider the chemical equation, <br/>`underset(68 g)(2H_(2)O_(2)) to 2H_(2)O + underset(22400 mL at N.T.P.)O_(2)` <br/>Now 68 g of `H_(2)O_(2)` give `O_(2)` at N.T.P. =22400 mL <br/>`therefore 1.7 g ` of `H_(2)O_(2)` will give `O_(2)=(22400)/(68)xx1.7=560` mL <br/> But 1.7 g of `H_(2)O_(2)` are present in 1000 mL of `H_(2)O_(2)` solution <br/>Hence, 1000 mL of `H_(2)O_(2)` solution gives 560 mL of `O_(2)` at N.T.P. <br/>`therefore ` 1 mL of `H_(2)O_(2)` solution will give `=(560)/(1000)=0.56 ` mL of `O_(2)` at N.T.P. <br/>`therefore` Volume strength of `H_(2)O_(2)` solution =0.56</body></html>


Discussion

No Comment Found

Related InterviewSolutions