Question

how many liters of a 60% antifreeze solution must be added to 8L of a 10% antifreeze solution to produce a 20% antifreeze solution?

Answer 1

Answer: 2Liters

Explanation:

The expression used will be :

$M_1V_1+M_2V_2=M_3V_3$

where,

$M_1$ = concentration of first antifreeze= 60%

$M_2$ = concentration of second  antifreeze= 10%

$V_1$ = volume of first antifreeze = x L

$V_2$ = volume of second antifreeze = 8 L

$M_3$ = concentration of final antifreeze solution= 20%

$V_3$ = volume of final antifreeze = (x+8) L

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

$60\times x+10\times 8=20\times (x+8)$

$x=2L$

Therefore, the volume of 60% antifreeze solution that must be added is 2L