Solved Dilution Problems Worksheet (MV M2V2) A How much of a ... - Free Printable

Here are the step-by-step solutions for each problem on the worksheet.

Dilution Problems Worksheet ($M_1V_1 = M_2V_2$)

1. How much of a 15.0 M stock solution do you need to prepare 250 ml of a 2.35 M HF solution?
* Identify variables:
* $M_1$ (Stock Molarity) = $15.0 \text{ M}$
* $V_1$ (Volume needed) = ?
* $M_2$ (Final Molarity) = $2.35 \text{ M}$
* $V_2$ (Final Volume) = $250 \text{ ml}$
* Formula: $M_1V_1 = M_2V_2$
* Rearrange: $V_1 = \frac{M_2V_2}{M_1}$
* Calculation: $V_1 = \frac{2.35 \times 250}{15.0} = \frac{587.5}{15.0} \approx 39.166...$
* Result: Round to 3 significant figures: 39.2 ml

2. If 455 ml of 6.0 M HNO$_3$ is diluted to 2.5 L, what is the molarity of the diluted solution?
* Unit Check: Convert volume to Liters so units match. $455 \text{ ml} = 0.455 \text{ L}$.
* Identify variables:
* $M_1 = 6.0 \text{ M}$
* $V_1 = 0.455 \text{ L}$
* $M_2 = ?$
* $V_2 = 2.5 \text{ L}$
* Formula: $M_1V_1 = M_2V_2$
* Rearrange: $M_2 = \frac{M_1V_1}{V_2}$
* Calculation: $M_2 = \frac{6.0 \times 0.455}{2.5} = \frac{2.73}{2.5} = 1.092$
* Result: Round to 2 significant figures (based on 6.0 and 2.5): 1.1 M

3. If 65.5 ml of HCl stock solution is used to make 450 ml of a 0.675 M HCl dilution, what is the molarity of the stock solution?
* Identify variables:
* $M_1 = ?$
* $V_1 = 65.5 \text{ ml}$
* $M_2 = 0.675 \text{ M}$
* $V_2 = 450 \text{ ml}$
* Formula: $M_1V_1 = M_2V_2$
* Rearrange: $M_1 = \frac{M_2V_2}{V_1}$
* Calculation: $M_1 = \frac{0.675 \times 450}{65.5} = \frac{303.75}{65.5} \approx 4.637...$
* Result: Round to 3 significant figures: 4.64 M

4. How do you prepare 500 ml of a 1.77 M H$_2$SO$_4$ solution from an 18.0 M H$_2$SO$_4$ stock solution?
* Step 1: Calculate the volume of stock needed.
* $M_1 = 18.0 \text{ M}$, $V_1 = ?$, $M_2 = 1.77 \text{ M}$, $V_2 = 500 \text{ ml}$
* $V_1 = \frac{1.77 \times 500}{18.0} = \frac{885}{18.0} = 49.166...$
* Round to 49.2 ml.
* Step 2: Describe the preparation.
* You need to measure out 49.2 ml of the concentrated acid and add water until the total volume reaches 500 ml.
* *(Note: Always add acid to water, not water to acid, for safety).*

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Extra Molarity Problems for Practice

5. How many moles of LiF would be required to produce a 2.5 M solution with a volume of 1.5 L?
* Formula: $\text{Moles} = \text{Molarity} \times \text{Volume (L)}$
* Calculation: $2.5 \text{ mol/L} \times 1.5 \text{ L} = 3.75 \text{ moles}$
* Result: 3.75 moles (or 3.8 moles if rounding to sig figs)

6. How many moles of Sr(NO$_3$)$_2$ would be used in the preparation of 2.50 L of a 3.5 M solution?
* Formula: $\text{Moles} = \text{Molarity} \times \text{Volume (L)}$
* Calculation: $3.5 \text{ mol/L} \times 2.50 \text{ L} = 8.75 \text{ moles}$
* Result: 8.75 moles (or 8.8 moles if rounding to sig figs)

7. What is the molarity of a 500-ml solution containing 249 g of KI?
* Step 1: Find Molar Mass of KI.
* K (Potassium) = $39.10 \text{ g/mol}$
* I (Iodine) = $126.90 \text{ g/mol}$
* Total = $166.00 \text{ g/mol}$
* Step 2: Convert grams to moles.
* $\text{Moles} = \frac{249 \text{ g}}{166.00 \text{ g/mol}} = 1.5 \text{ moles}$
* Step 3: Convert volume to Liters.
* $500 \text{ ml} = 0.500 \text{ L}$
* Step 4: Calculate Molarity.
* $M = \frac{1.5 \text{ moles}}{0.500 \text{ L}} = 3.0 \text{ M}$
* Result: 3.0 M

8. How many grams of CaCl$_2$ would be required to produce a 3.5 M solution with a volume of 2.0 L?
* Step 1: Calculate moles needed.
* $\text{Moles} = 3.5 \text{ M} \times 2.0 \text{ L} = 7.0 \text{ moles}$
* Step 2: Find Molar Mass of CaCl$_2$.
* Ca (Calcium) = $40.08 \text{ g/mol}$
* Cl (Chlorine) = $35.45 \text{ g/mol} \times 2 = 70.90 \text{ g/mol}$
* Total = $110.98 \text{ g/mol}$
* Step 3: Convert moles to grams.
* $\text{Mass} = 7.0 \text{ moles} \times 110.98 \text{ g/mol} = 776.86 \text{ g}$
* Result: Round to 2 significant figures: 780 g (or 777 g if keeping more precision).

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Final Answer:
1. 39.2 ml
2. 1.1 M
3. 4.64 M
4. Measure 49.2 ml of the 18.0 M stock solution and dilute it with water to a total volume of 500 ml.
5. 3.75 moles
6. 8.75 moles
7. 3.0 M
8. 780 g