Question

If 0.1718 l of a 0.3556-m c3h7oh solution is diluted to a concentration of 0.1222 m, what is the volume of the resulting solution

Answer 1

Answer : The volume is found to be 0.4999 L.

Explanation : In the given case, the two concentrations of the solution are given and the volume concentration given for one and other needs to be identified.

So, according to the formula :-  $m _{1}V_{1}=m_{2} V_{2}$

Here suppose the  $V _{1}$ is 0.1718 L moles and $m _{1}$  as 0.3556 moles, then we need to determine the $V _{2}$  as $m _{2}$  is mentioned as 0.1222 moles.

Therefore from the formula we get,   $m _{1}V_{1}=m_{2} V_{2}$ 

 So, $V_{2}$ = (0.1718 X 0.3556) / 0.1222 = 0.4999 L

Answer 2

Answer:  0.5 Liters

Explanation :

The dilution law is:

$M_1V_1=M_2V_2$

where,

$M_1$ = concentration of stock solution = 0.3556 M

$M_2$ = concentration of resulting solution  = 0.1222 M

$V_1$ = volume of stock solution= 0.1718 L

$V_2$ = volume of resulting solution  = ?

Now put all the given values in the above law, we get the volume of the resulting solution.

$(0.3556\times 0.1718=(0.1222\times V_2)$

$V_2=0.5L$

Thus the volume of the resulting solution is 0.5 Liters.