1.

25 ml of hydrogen peroxide solution were added to excess of acidified potassiutm iodide solution. The iodine so liberated required 20mL of 0.1 sodium this sulphate solution. Calculate the strenth in terms of normality, percentage and volume.

Answer» <html><body><p></p>Solution :Let the molarity of `H_(2)O_(2)=N_(x)` <br/> 20mL of 0.1 N `Na_(2)SO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)-=20mL of 0.1 I_(2)=25mL.N_(x).H_(2)O_(2)` <br/> Normality of `H_(2)O_(2)(N_(x))=(20xx0.1)/(25)=0.08N` <br/> <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)`=Normality`xx`Equivalent mass=`0.08xx17=1.36gL^(-1)` <br/> Percentage of `H_(2)O_(2)=(1.36)/(1000)xx100=0.136` <br/> This means that 0.136g of `H_(2)O_(2)` are <a href="https://interviewquestions.tuteehub.com/tag/present-1163722" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENT">PRESENT</a> in 100mL solution or 0.00136g of `H_(2)O_(2)` are present in <a href="https://interviewquestions.tuteehub.com/tag/1ml-1803748" style="font-weight:bold;" target="_blank" title="Click to know more about 1ML">1ML</a> of the solution <br/> Now, 68g of `H_(2)O_(2)` give oxygen at N.T.P. =22400mL <br/> 0.00136g of `H_(2)O_(2)` given oxygen at N.T.P.`=(22400)/(68)xx0.00136=0.448mL` <br/> Thus. 1mL of `H_(2)O_(2)` givwes 0.048mL of oxgyen at N.T.P.</body></html>


Discussion

No Comment Found

Related InterviewSolutions