How many liters (L) of a $10 \%$ silver iodide solution must be mixed with $5 \mathrm{~L}$ of a $4 \%$ silver iodide solution to get a $6 \%$ solution?
A) $3.5 \mathrm{~L}$
B) $2.5 \mathrm{~L}$
C) $5.0 \mathrm{~L}$
D) $1.5 \mathrm{~L}$

Question

Quizplus · Accepted Answer

To find the correct amount, we can use the equation for mixture problems: $C_1V_1 + C_2V_2 = C_{final}(V_1 + V_2)$, where $C_1$ and $C_2$ are the concentrations of the initial solutions, $V_1$ and $V_2$ are the volumes of the initial solutions, and $C_{final}$ is the concentration of the final solution. Plugging in the given values: $0.10V_1 + 0.04(5) = 0.06(V_1 + 5)$. Solving for $V_1$ gives $V_1 = 2.5$ 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