Question

how many ml of water are needed to dilute 65ml 7m KCl
to 2M​

Answer 1

Initial Concentration = 7M

Final Concentration = 2M

Let the final volume of solution be V_(2).

So, we can say :

$V_{1} \times S_{1} = V_{2} \times S_{2}$

$\implies 65 \times 7 = V_{2} \times 2$

$\implies V_{2} = \dfrac{65 \times 7}{2}$

$\implies V_{2} = 227.5 \: ml$

So, amount of water to be added :

$= 227.5 - 65 = 162.5 \: ml$

So, 162.5 mL of water has to be added.

Answer 2

Explanation:

Initial Concentration = 7M

Final Concentration = 2M

Let the final volume of solution be V_(2).

So, we can say :

V_{1} \times S_{1} = V_{2} \times S_{2}V

1

×S

1

=V

2

×S

2

\implies 65 \times 7 = V_{2} \times 2⟹65×7=V

2

×2

\implies V_{2} = \dfrac{65 \times 7}{2} ⟹V

2

=

2

65×7

\implies V_{2} = 227.5 \: ml⟹V

2

=227.5ml

So, amount of water to be added :

= 227.5 - 65 = 162.5 \: ml=227.5−65=162.5ml

So, 162.5 mL of water has to be added.