Question

to prepare 0.1 M kmno4 solution in 250 ml flask the weight of kmno4 required​

Answer 1

Answer:

Molarity = (mass/molar mass) × 1000/volume(in ml)

0.1 = (mass/158) × 1000/250

Mass = 0.1 × 158 / 4

Mass = 3.95 g

Mass required is 3.95 g

Pls follow

Answer 2

The weight of $KMnO_4$ required​ is 3.95 gram .

We need to prepare 0.1 M $KMnO_4$ solution in 250 ml flask .

We need to find the weight of $KMnO_4$ required .

Let , weight required be w .

Now , we know Molarity , $M=\dfrac{n\times1000}{V(in\ ml)}$   .......(1)

Here n is number of moles given by , $n=\dfrac{weight}{molecular\ mass}=\dfrac{w}{M.M}$    ......(2)

Also , molecular mass of $KMnO_4$ is 158.034 gm/mol .

Putting value of M.M in equation 1 we get :

$M=\dfrac{n\times1000}{V(in\ ml)}\\\\0.1=\dfrac{\dfrac{w}{158.034}\times 1000}{250}\\\\w=3.95\ gm$

Hence , this is the required solution .

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Volumetric Analysis

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