Respuesta :
Explanation:
(a) As we know that relation between normality and volume of two solutions is as follows.
[tex]N_{1}V_{1} = N_{2}V_{2}[/tex]
As the given data is as follows.
[tex]N_{1}[/tex] = ?, [tex]N_{2}[/tex] = 0.050 N
[tex]V_{1}[/tex] = 20 ml, [tex]V_{2}[/tex] = 16.25 ml
Therefore, putting the given values into the above formula as follows.
[tex]N_{1}V_{1} = N_{2}V_{2}[/tex]
[tex]N_{1} \times 20 ml = 0.050 N \times 16.25 ml[/tex]
[tex]N_{1}[/tex] = 0.0406 N
Hence, normality of solution A is 0.0406 N.
(b) As gram equivalent weight of KBr is 119 g/mol.
Formula to calculate normality is as follows.
Normality = [tex]\frac{weight}{gm equi. wt} \times \frac{1000}{V_{ml}}[/tex]
As volume is 200 ml, normality is 0.0406 N, and gram equi. wt is 119 g/mol. Therefore, putting these values into the above formula as follows.
Normality = [tex]\frac{weight}{gm equi. wt} \times \frac{1000}{V_{ml}}[/tex]
0.0406 N = [tex]\frac{weight}{119 g/mol} \times \frac{1000}{200 ml}[/tex]
weight = 0.9668 g
As 0.9668 g of KBr is present in 1.1250 g sample. Hence, weight percentage will be calculated as follows.
Weight % of KBr = [tex]\frac{0.9668 g}{1.1250 g}[/tex]
= 85.94%
Thus, we can conclude that C% w/w of KBr in the original sample is 85.94%.
