19(100 * 0.1) = ( 0.25 * v)/2
=>20/0.25 =v =80ml
100 ml of 0.1 M NaAl(OH)2CO3 is neutralised by 0.25 N Hcl to form Nacl ,Alcl3 & CO2
Volume of Hcl reqd is ...
(A) 10 mL
(B) 40 mL
(C) 100 mL
(D) 160 mL
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9 Answers
24the main prob i suppose will be in judging eq wt of NaAl(OH)2CO3
which is 1
becoz Na+ and (Al3+(OH-)2CO32-)-
*edit :its not eq wt as i wrote...its n factor *
19if in a balanced chemical equation , we find x* acid + y* base gives the correct balance, then
V1M1 / X = V2M2 / Y
(M-MOLARITY )
1answer should be 160ml naa!!!!
n-factor of given compound is 4
so by eq.concept
100*0.1*4=0.25*V
V=160ml