Question

how many grams of koh (80% pure) required to prepare 10L of 0.5 N solution of koh (in gm)​​

Answer 1

Explanation:

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Answer 2

To prepare 10L of 0.5 N solution of KOH 350 g KOH (80% pure) is required.

Given:

KOH in the 10L of 0.5 N solution is 80% pure.

To Find:

The weight of KOH that is required in the solution.

Solution:

According to the formula of Normality,

Normality= Number of gram equivalents × [volume of solution in litres]-1

Hence, the number of gram equivalents = 0.5 x 10 = 5

Also, the number of gram equivalents = $\frac{weight of solute}{Equivalent weight}$   (Equivalent weight of KOH= 39+16+1= 56 g)

$5= \frac{weight of solute}{56}$

Weight of solute= 5 x 56 = 280g

KOH is 80% pure.

So, Mass required x $\frac{80}{100}$ = 280 g

Mass required = 280×$\frac{100}{80}$= 350g of KOH.

Hence, 350 g KOH (80% pure) is required to prepare 10L of 0.5 N solution of KOH.

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