Question

Q1.0.63 gram oxalic acid (equivalent weight = 63) is dissolved in 250 ml
of solution. Find out the normality of solution.​

Answer 1

Explanation:

Given:

0.63 gram oxalic acid (equivalent weight = 63 g) is dissolved in 250 ml of solution.

To find:

Normality of solution?

Calculation:

Let normality be N :

$\rm \: N = molarity \times molar \: mass \div equivalent \: weightN$

$\rm \implies\: N = \dfrac{moles}{volume} \times molar \: mass \div equivalent \: weight$

$\rm \implies\: N = \dfrac{ (\frac{0.63}{126}) }{( \frac{250}{1000}) } \times 126 \div 63$

$N = 0.63 \times 4 \div 63$

$</p><p></p><p></p><p>\rm \implies\: N = 0.04 \: Normal$

So, normality of solution is 0.04 N.

Answer 2

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