Question

A sample containing KBr (M 119&/mol) and KNO3 (M 101 g/mol) was analysed by using the Mohr method, A 1.1250-g sample is dissolved into 200.0 ml solution (Solution A). 20.00-ml. of solution A was titrated with 16.25 ml of 0.OSON AgNOs (M 170) solution. a) Report C of ca solution & according to the tination data b) Calculate C% w/w of KBr in the original sample.

Answer 1

Step-by-step explanation:

(a) As we know that relation between normality and volume of two solutions is as follows.

`N_(1)V_(1) = N_(2)V_(2)`

As the given data is as follows.

`N_(1)` = ?,
`N_(2)` = 0.050 N

`V_(1)` = 20 ml,
`V_(2)` = 16.25 ml

Therefore, putting the given values into the above formula as follows.

`N_(1)V_(1) = N_(2)V_(2)`

`N_(1) * 20 ml = 0.050 N * 16.25 ml`

`N_(1)` = 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 =
`(weight)/(gm equi. wt) * (1000)/(V_(ml))`

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 =
`(weight)/(gm equi. wt) * (1000)/(V_(ml))`

0.0406 N =
`(weight)/(119 g/mol) * (1000)/(200 ml)`

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 =
`(0.9668 g)/(1.1250 g)`

= 85.94%

Thus, we can conclude that C% w/w of KBr in the original sample is 85.94%.