Question

A student prepared a stock solution by dissolving 20.0 g of koh in enough water to make 150. ml of solution. she then took 15.0 ml of the stock solution and diluted it with enough water to make 65.0 ml of a final solution. what is the concentration of koh for the final solution?

Answer 1

So first find moles of KOH
20/56 = 0.357

Find concentraion in 150ml
0.357 x 1000 / 150 = 2.38M

FInd concentration in diluted 65ml
Added Vol x concentraion / Total volume
15 x 2.38 / 65 = 0.579M

hope that helps 

Answer 2

The concentration of KOH for the final solution will be "0.549 M".

According to the question,

→ The moles of KOH will be:

= $\frac{20}{56}$

= $0.357$

→ In 150 ml, the final concentration will be:

= $0.357\times \frac{1000}{150}$

= $0.0238\times 100$

= $2.38 \ M$

hence,

→ In diluted 65 ml, the final concentration will be:

= $Added \ Volume\times \frac{Concentration}{Total \ Volume}$

By substituting the values, we get

= $15\times \frac{2.38}{65}$

= $0.2307\times 2.38$

= $0.549 \ M$

Thus the response above is correct.

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