Question

How many grams of divisive acid(molecular wt. 200) should be present in 100ml of its aqueous solution to give decinormal strength?

Answer 1

Here is your answer
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▶The molar mass of the acid is 200 g/mol

▶The volume of the acid solution is 0.1 L

▶The normality of the acid solution is 0.1 N


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The concentration of the dibasic acid solution is
c(acid) = n(acid)/V(acid)



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▶ n(acid) = c(acid) * V(acid)

▶ n(acid) = m(acid)/M(acid)


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=> mass of acid will be

= m(acid) = n(acid) * M(acid)

= c(acid) * V(acid) *M(acid)

= N(acid) * f(eq) * V(acid) * M(acid)

= 0.1 N * 0.5 * 0.1 L * 200 g/mol

= 1 g

Answer 2

Hey User!
Here's ur answer!

⏩ Given Volume = 100 mL = 0.1 L.

⏩ Equivalent weight of the given acid = 200/2 = 100 g/mol.

⏩ Strength of the given acid = 0.1 N.

⏩ Mass = Volume.Eq.weight.Strength/1000.

⏩ Mass = 100×100×0.1/1000.

⏩ Mass = 1 g.
Hope it helps!!