Weight of KMnO4 required to prepare 250 ml of its (N/10) . solution if equivalent weight of KMnO4 is 31.6, will be:

Question

Weight of KMnO4 required to prepare 250 ml of its (N/10) . solution if equivalent weight of KMnO4 is 31.6, will be: (A)  0.64g  (B)  1.24g  (C)  1.78g  (D)  0.79g

Tardigrade Admin · Accepted Answer

Milli equivalents of KMnO4 solution =(1/10)× 250=25 Equivalents of KMnO4=(25/1000)=0.025 Thus, required weight = number of equivalents × equivalent weight. =0.025× 31.6 =0.79g

Q. Weight of $K M n O_{4}$ required to prepare 250 ml of its $\frac{N}{10}$ . solution if equivalent weight of $K M n O_{4}$ is 31.6, will be:

$0.64 g$

$1.24 g$

$1.78 g$

$0.79 g$

Milli equivalents of $K M n O_{4}$ solution $= \frac{1}{10} \times 250 = 25$ Equivalents of $K M n O_{4} = \frac{25}{1000} = 0.025$ Thus, required weight = number of equivalents $\times$ equivalent weight. $= 0.025 \times 31.6$ $= 0.79 g$

Q. Weight of KMnO4​ required to prepare 250 ml of its 10N​ . solution if equivalent weight of KMnO4​ is 31.6, will be:

0.64g

1.24g

1.78g

0.79g