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5 g of a sample of brass were dissolved in 1 litre dil. H_(2)SO_(4). 20 mL of this solution were mixed with KI and liberated iodine required 20 mL of 0.0327 N hypo solution for titration. Calculate the amount of copper in the alloy.

Answer» <html><body><p></p>Solution :When brass is extracted with concentrated `H_(2)SO_(4)`, it <a href="https://interviewquestions.tuteehub.com/tag/gives-1007647" style="font-weight:bold;" target="_blank" title="Click to know more about GIVES">GIVES</a> copper sulphate. <br/> `2[Cu+2H_(2)SO_(4)to CuSO_(4)+SO_(2)+2H_(2)O]` <br/> `2CuSO_(4)+4KI to 2K_(2)SO_(4)+2CuI_(2)` <br/> `2CuI_(2) to Cu_(2)I_(2)+I_(2)` <br/> `2Na_(2)S_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)+I_(2) to 2NaI+Na_(2)S_(4)O_(6)` <br/> 2moles `Cu -= 1 " mole" I_(2) 0-2` mole <a href="https://interviewquestions.tuteehub.com/tag/hypo-1034629" style="font-weight:bold;" target="_blank" title="Click to know more about HYPO">HYPO</a> <br/> 20 mL of solution reacts with 20 mL of 0.0327 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> hypo <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> 1000 mL` of solution will react with 1000 mL of 0.0327 N hypo <br/> No. of moles of hypo used `=("Mass")/("Molecular mass" (158))` <br/> `=(Exx NxxV)/(1000xx158)` <br/> where, E=158, N=0.327 given, V=1000 mL <br/> `therefore` No. of moles of hypo used `=(158xx0.0327xx1000)/(1000xx158)=0.0327` <br/> No. of moles of Cu=No. of moles of hypo =0.0327 mole <br/> Mass of copper in brass `=0.0327xx63.5=2.07645` <br/> % of copper in brass `=(2.07645)/(5)xx100=41.529%`</body></html>


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