Question

Calculate the normality of a solution obtained by mixing 0.01 l of N/10 NaOH and 40 ml of seminormal KOH solution. :-) ;-)

Answer 1

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Given Normality of KOH= 0.1 N,

Volume of HCl = 100 ml

Normality of NaOH = 0.2 N,

Volume of NaOH = 100 ml

∴ Miliequivalents of KOH =0.1×100=10

∴ Miliequivalents of NaOH =0.2×200=20

KOH+NaOH→NaCl+H2o

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Miliquivalents before reaction 10 20

Milliequivalents after reaction 0 10 10

No. of moles left in solution = 20 - 10 = 10

Normality of resulting solution = $\frac{total volume of solution}{no .of moles left in solution}$

No. of moles left in solution

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= $\frac{10}{100+100}$

= $\frac{10}{200}$

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=0.05N. .... ANSWER

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Answer 2

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To make 0.1N NaOH solution = dissolve 40 grams of NaOH in 1L of water. For 100 ml of water = (4/1000) × 100 = 0.4 g of NaOH. Thus, the amount of NaOH required to prepare 100ml of 0.1N NaOH solution is 0.4 g of NaOH