How many liters (L) of a $10 \%$ silver iodide solution must be mixed with $6 \mathrm{~L}$ of a $4 \%$ silver iodide solution to get a $6 \%$ solution?
A) $6.0 \mathrm{~L}$
B) $3.0 \mathrm{~L}$
C) $4.0 \mathrm{~L}$
D) $2.0 \mathrm{~L}$

Question

Quizplus · Accepted Answer

To solve this, we use the equation for mixing solutions: $(\text{concentration of first solution}) \times (\text{volume of first solution}) + (\text{concentration of second solution}) \times (\text{volume of second solution}) = (\text{final concentration}) \times (\text{total volume})$. Let $x$ be the volume of the $10\%$ solution. The equation becomes $0.10x + 0.04(6) = 0.06(x + 6)$. Solving for $x$ gives $x = 3.0$ liters.

How Many Liters (L) of A 10%10 \%10% Silver Iodide Solution Must Be Mixed With

How Many Liters (L) of A $10 \%$ Silver Iodide Solution Must Be Mixed With