Question

4.2g KOH in 100 ml HCL solution

Answer 1

Answer:

Let mass of KOH be present in mixture =a g

and Mass of Ca(OH)2​=(4.2−a)g

Eq. mass of KOH=56; Eq. mass of Ca(OH)2​=274​=37

g equivalent of KOH+g equivalent of Ca(OH)2​=g equivalent of the acid

56a​+37(4.2−a)​=0.1

or 37a−56a=0.1×56×37−4.2×56

or 19a=28

a=1928​=1.47

Mass of KOH in the sample =1.47 g

Percentage of KOH=35

and Percentage of Ca(OH)2​=100−35=65

Explanation:

Answer 2

2g koh in 100 ml hcl solution