Question

question 12 of 12 > a 47.0 ml aliquot of a 0.600 m stock solution must be diluted to 0.100 m. assuming the volumes are additive, how much water should be added?

Explanation:

Step1: Use dilution formula

$M_1V_1 = M_2V_2$, where $M_1 = 0.600\ M$, $V_1=47.0\ mL$, $M_2 = 0.100\ M$.
We can find $V_2$ as $V_2=\frac{M_1V_1}{M_2}$.

Step2: Calculate $V_2$

$V_2=\frac{0.600\ M\times47.0\ mL}{0.100\ M}=282\ mL$.

Step3: Find volume of water to add

The volume of water to add $V_{water}=V_2 - V_1$.
$V_{water}=282\ mL - 47.0\ mL = 235\ mL$.

Answer:

$235\ mL$