Question

dilutions worksheet - solutions

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? 0.19 m (the final volume is 900 ml, set up the equation from that)
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? 0.033 m (the final volume is 750 ml, set up the equation from that. note that the phrasing difference between problems 1 and 2 makes a big difference in the final answer).
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? 0.63 m (this is the opposite of a dilutions problem - the v₂ value is smaller than v₁, but otherwise the equation is no different.)
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? 1500 ml

Explanation:

Step1: Use the dilution formula

The dilution formula is \(M_1V_1 = M_2V_2\). For problem 3, \(M_1 = 0.50\space M\), \(V_1=750\space mL\), and \(V_2 = 750 - 150=600\space mL\). We need to find \(M_2\).

Step2: Rearrange the formula to solve for \(M_2\)

From \(M_1V_1 = M_2V_2\), we can get \(M_2=\frac{M_1V_1}{V_2}\). Substitute the values: \(M_2=\frac{0.50\times750}{600}\)

$$M_2=\frac{375}{600}= 0.625\approx0.63\space M$$

For problem 4, we use the same formula \(M_1V_1 = M_2V_2\). Here \(M_1 = 0.63\space M\) (from problem 3 result, more accurately \(M_1=\frac{0.5\times750}{600}\)), \(M_2 = 0.25\space M\), and we need to find \(V_2\). Rearranging gives \(V_2=\frac{M_1V_1}{M_2}\). Substitute \(M_1=\frac{0.5\times750}{600}\), \(V_1 = 600\space mL\) (after evaporation in problem 3)

$$V_2=\frac{\frac{0.5\times750}{600}\times600}{0.25}=\frac{0.5\times750}{0.25}=1500\space mL$$

Answer:

1. \(0.19\space M\)
2. \(0.033\space M\)
3. \(0.63\space M\)
4. \(1500\space mL\)