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? 13. a concentrated sulfuric acid solution is available in a lab. a 100 - ml solution with a concentration of 0.75 mol/l is required. write out the pre - lab and procedure for making this solution.

Explanation:

Step1: Calculate moles before dilution

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

Step2: Calculate final volume after dilution

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

Step3: Calculate volume of water to add

The volume of water $V_{water}$ to be added is the difference between the final volume and the initial volume. $V_{water}=V_2 - V_1$.
$V_{water}=157.5\ mL- 105\ mL = 52.5\ mL$

Answer:

$52.5\ mL$