First slide

Concentration methods

Question

24.5 grams of Sulphuric acid is dissolved in a one litre solution. Normality of the solution is…………N

Moderate

A

1
B

0.5
C

0.25
D

2

Solution

$N = Number of gram equivalents \times \frac{1}{V (l)}; N = Number of gram equivalents \times \frac{1000}{V (ml)}; Number of gram equivalents = \frac{w_{solute}}{GEW of solute}; GEW = \frac{GMW or GFW}{valency}; GEW of H_{2} {SO}_{4} = \frac{98}{2} g; N = \frac{24.5}{49} \times \frac{1}{1} = 0.5 N;$

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