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Solutions

Question

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The normality of 10% (w/v) of acetic acid is

1 N

1.3 N

1.7 N

1.9 N

Solution

C.

1.7 N

Normality is given as,

$\frac{W x 1000}{m x V (mL)} Here, m = mass of acetic acid = \frac{10 x 1000}{60 x 100} = 1.66 ∴ N = 1.66 N N \approx 1.7 N$

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