JEE MAIN - Chemistry (2021 - 25th February Evening Shift - No. 19)
Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained.
(i) 4.5 mL
(ii) 4.5 mL
(iii) 4.4 mL
(iv) 4.4 mL
(v) 4.4 mL
If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is ________ M. (Rounded off to the nearest integer)
(i) 4.5 mL
(ii) 4.5 mL
(iii) 4.4 mL
(iv) 4.4 mL
(v) 4.4 mL
If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is ________ M. (Rounded off to the nearest integer)
Answer
6
Explanation
Average burette reading = Volume of NaOH solution (V1)
$$ = {{4.5 + 4.5 + 4.4 + 4.4 + 4.4} \over 5}$$
= 4.44 mL
Strength of NaOH solution = S1(M) (say) = S1(N)
Volume of oxalic acid solution (V2) = 10 mL
Strength of oxalic acid solution (S2) = 1.25 M = 1.25 $$\times$$ 2 N
So, $${V_1}{S_1} = {V_2}{S_2}$$ ($$\because$$ Law of equivalence)
$$ \Rightarrow {S_1} = {{{V_2}{S_2}} \over {{V_1}}} = {{10 \times (1.25 \times 2)} \over {4.44}} = 5.63$$ N
$$ \simeq 6M = 6M$$
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