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A sample of coconut oil weighing 1.5763 g is mixed with 25 mL of 0.4210 M KOH. Some KOH is used in saponification of coconut oil. After the saponification is complete, 8.46 mL of 0.2732 M H_(2)SO_(4) is required to neutralize excess KOH. The saponification number of peanut oil is :

Answer» <html><body><p>209.6<br/>98.9<br/>108.9<br/>218.9</p>Solution :Number of <a href="https://interviewquestions.tuteehub.com/tag/milliequivalent-560828" style="font-weight:bold;" target="_blank" title="Click to know more about MILLIEQUIVALENT">MILLIEQUIVALENT</a> of KOH added`=25xx0.421=10.525` <br/> Number of milliequivalents <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> unreacted <br/> =Number of milliequivalents of `H_(2)SO_(4)` used <br/> `= 8.46xx0.2732xx24.623 "" ("Here, basicity of " H_(2)SO_(4)=2)` <br/> Number of milliequivalents of KOH used by oil =10.525-4.623=5.902 <br/> <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of KOH used `=(5.902xx56)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)=0.3305 g =330.5 <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>` <br/> Saponification number = Mass of KOH in mg used by 1 g oil or fat <br/> `=(0.3305xx1000)/(1.5763)=209.6`</body></html>


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