Question

The molecular weight of KMnO4 is 158 g/mol. If you wanted to make 500 mL of a 0.1M stock solution of KMnO4, how much solid KMnO4 would you need to add before filling mL flask of water?

Answer 1

Solid KMnO₄ needed = 7.9 g

Further explanation

Given

MW KMnO₄ = 158 g/mol

500 mL(0.5 L) of a 0.1M stock solution of KMnO₄

Required

solid KMnO₄

Solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

$\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

Input the value :

n = M x V

n = 0.1 M x 0.5 L

n = 0.05 mol

Mass KMnO₄ :

= mol x MW

= 0.05 x 158 g/mol

= 7.9 g

Answer 2

The solid KMnO₄ a person need to add before filling water flask will be:

"7.9 g".

Molecular weight

According to the question,

Molecular weight of KMnO₄ = 158 g/mol

Volume of solution = 500 mL or,

                                = 0.5 L

Molarity = 0.1 M

We know the relation,

Molarity, M = $\frac{n}{V}$

or,

→ n = M × V

By substituting the values,

     = 0.1 × 0.5

     = 0.05 mol

hence,

The mass of KMnO₄ will be:

= mol × Molecular weight

= 0.05 × 158

= 7.9 g

Thus the response above is appropriate.

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