Question

You need a 30% alcohol solution. On hand, you have a 200 mL of a 10% alcohol mixture. You also have 55% alcohol mixture. How much of the 55% mixture will you need to add to obtain the desired solution?

Answer 1

Answer: 160 ml

Explanation:

The expression used will be :

$C_1V_1+C_2V_2=C_3V_3$

where,

$C_1$ = concentration of Ist alcohol solution= 10%

$C_2$ = concentration of 2nd alcohol solution= 55%

$V_1$ = volume of Ist alcohol solution = 200 ml

$V_2$ = volume of  2nd alcohol solution= v ml

$C_3$ = concentration of resulting alcohol solution= 30%

$V_2$ = volume of resulting alcohol solution= (v+200) ml

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

$(10\times 200)+(55\times v)=(30\times (v+200)ml)$

By solving the terms, we get :

$v=160ml$

Therefore, the volume of 55% mixture  needed to be added to obtain the desired solution is 160 ml.