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HNO(3) used as a reagent has specific gr...

`HNO_(3)` used as a reagent has specific gravity of `1.42 g mL^(-1)` and contains 70% by strength `HNO_(3)`. Calcualte:
a. Normality of acid

Text Solution

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The correct Answer is:
A, B, C
a. `(Mw = Ew of HNO_(3) = 1 + 14 + 16 xx 3 = 63 g)`
`N = (% "by Weight" xx 10 xx d)/(Ew) = (70 xx 10 xx 1.42)/(63) = 15.78`
b. `70 xx 1.42 g` of pure acid is present in `100 mL`.
`63 g` of pure acid is present in `= (100 xx 63)/(70 xx 1.42)`
`= 63.38 mL`
c. mEq of `HNO_(3)` (before) = mEq of dil. `HNO_(3)`
(after dilution)
(`:'` mEq does not change on dilution)
`N_(1) V_(1) = N_(2) V_(2)`
`15.78 xx 2 = 1 xx v_(2)`
Volume of `H_(2) O` added `= 31.56 - 2 = 29.56 mL`
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