Question

1. what volume of distilled water must be added to 105 ml of 0.600 - mol/l sulfuric acid solution in order to dilute it to a concentration of 0.400 mol/l?

Explanation:

Step1: Calculate initial moles of sulfuric acid

Use the formula $n = cV$, where $n$ is the number of moles, $c$ is the concentration and $V$ is the volume. Given $c_1=0.600\ mol/L$ and $V_1 = 105\ mL=0.105\ L$. So $n = c_1V_1=0.600\ mol/L\times0.105\ L = 0.063\ mol$.

Step2: Calculate the final volume of the diluted solution

Since the number of moles remains the same during dilution, and the final concentration $c_2 = 0.400\ mol/L$. Using the formula $V=\frac{n}{c}$, we have $V_2=\frac{n}{c_2}=\frac{0.063\ mol}{0.400\ mol/L}=0.1575\ L = 157.5\ mL$.

Step3: Calculate the volume of water added

The volume of water added $V_{water}=V_2 - V_1$. Substitute $V_2 = 157.5\ mL$ and $V_1=105\ mL$ into the formula. So $V_{water}=157.5\ mL - 105\ mL = 52.5\ mL$.

Answer:

52.5 mL