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Q.

Equivalent Weight of dil.HNO₃ when it reacts with Zn is (At.wt of Zn = 65)

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a

$\frac{65}{3}$

b

65

c

$\frac{65\times 4}{3}$

d

$\frac{65\times 5}{4}$

answer is D.

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Detailed Solution

4Zn+10 HNO₃(dil.) $\to$ 4Zn(NO₃)₂+5H₂O+N₂O

HNO₃ in this reaction acts like both acid and Oxidant

E.W of HNO₃ as an acid. It forms Zn(NO₃)₂

No.of replaceable H-atoms in HNO₃ = 1

E.W of HNO₃ (acid) = M/1 =M

E.W of HNO₃ as an Oxidant

\mathop N\limits^{ + 5} O_3^ - \to \mathop {{N_2}}\limits^{ + 1} O

decrease in Ox. state =4

E.W of HNO₃ (oxidant) = M/4

\therefore Over all \;EW\; of\; HNO3 =\frac{M}{1}+\frac{M}{4}=\frac{5M}{4}

=\frac{5(65)}{4}

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