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1.16 g CH3(CH2)nCOOH was burnt in excess air & the resultant gases ( CO2 & H2O )were passed thru excess NaOH solution .
The resulting solution is divided into 2 equal parts.One part required 50 mL of 1 N Hcl for neutralization using phenolphthalein as indicator. Another part required 80 mL of 1 N Hcl for neutralization using methyl orange as indicator.
Find the value of n?
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10 Answers
1357CH3(CH2)nCOOH→(n+2)CO2
mill moles of CH3(CH2)nCOOH=1.16*1000/(12*(n+2)+2n+4+32)=1.16*1000/(14*(n+2)+32)
millimoles of CO2=1.16*1000/(14+32/(n+2))
CO2+NaOH=Na2CO3
millimoles of CO2=milllimoles of Na2CO3=x
millimoles of NaOH left=y
x/2+y=50
x+y=80
so x=60
30=1.16*1000/(14+32/(n+2))
14+32/(n+2)=1160/60=58/3
32/(n+2)=58/3-14=16/3
n+2=6
n=4
