Question

dilutions worksheet

1. if i have 340 ml of a 0.5 m nabr solution, what will the concentration be if i add 560 ml more water to it?
2. if i dilute 250 ml of 0.10 m lithium acetate solution to a volume of 750 ml, what will the concentration of this solution be?
3. if i leave 750 ml of 0.50 m sodium chloride solution uncovered on a windowsill and 150 ml of the solvent evaporates, what will the new concentration of the sodium chloride solution be?
4. to what volume would i need to add water to the evaporated solution in problem 3 to get a solution with a concentration of 0.25 m?

Explanation:

Step1: <Identify the formula>

The formula for dilution is \(C_1V_1 = C_2V_2\), where \(C_1\) and \(V_1\) are the initial concentration and volume, and \(C_2\) and \(V_2\) are the final concentration and volume.

Step2: <Solve for problem 1>

Given \(C_1 = 0.5\space M\), \(V_1=340\space mL\), \(V_2=(340 + 560)\space mL = 900\space mL\).
Using \(C_1V_1 = C_2V_2\), we have \(C_2=\frac{C_1V_1}{V_2}\).
Substitute the values: \(C_2=\frac{0.5\times340}{900}=\frac{170}{900}\approx0.19\space M\).

Step3: <Solve for problem 2>

Given \(C_1 = 0.10\space M\), \(V_1 = 250\space mL\), \(V_2=750\space mL\).
Using \(C_1V_1 = C_2V_2\), \(C_2=\frac{C_1V_1}{V_2}\).
Substitute the values: \(C_2=\frac{0.10\times250}{750}=\frac{25}{750}\approx0.033\space M\).

Step4: <Solve for problem 3>

Given \(C_1 = 0.50\space M\), \(V_1 = 750\space mL\), \(V_2=(750 - 150)\space mL=600\space mL\).
Using \(C_1V_1 = C_2V_2\), \(C_2=\frac{C_1V_1}{V_2}\).
Substitute the values: \(C_2=\frac{0.50\times750}{600}=\frac{375}{600}=0.625\space M\).

Step5: <Solve for problem 4>

From problem 3, \(C_1 = 0.625\space M\), \(V_1 = 600\space mL\), \(C_2 = 0.25\space M\).
Using \(C_1V_1 = C_2V_2\), \(V_2=\frac{C_1V_1}{C_2}\).
Substitute the values: \(V_2=\frac{0.625\times600}{0.25}=\frac{375}{0.25}=1500\space mL\).

Answer:

1. \(0.19\space M\)
2. \(0.033\space M\)
3. \(0.625\space M\)
4. \(1500\space mL\)