Question

molarity

1. what is the molarity of a solution in which 58g of nacl are dissolved in 1.0l of solution?
2. what is the molarity of a solution in which 10.0g agno₃ is dissolved in 500ml of solution?
3. how many grams of kno₃ should be used to prepare 2.00l of a 0.500m solution?
4. to what volume should 5.0g of kcl be diluted in order to prepare a 0.25m solution?
5. how many grams of cuso₄·5h₂o are needed to prepare 100ml of a 0.10m solution?
6. what is the molarity of a solution in which 0.65g of ki was dissolved to make 0.300 l of solution?

Explanation:

Step1: Find molar mass of NaCl

Molar mass of $\text{NaCl} = 23 + 35.5 = 58.5\ \text{g/mol}$

Step2: Calculate moles of NaCl

$n = \frac{m}{M} = \frac{58\ \text{g}}{58.5\ \text{g/mol}} \approx 0.991\ \text{mol}$

Step3: Calculate molarity (M)

$M = \frac{n}{V} = \frac{0.991\ \text{mol}}{1.0\ \text{L}} \approx 1.0\ \text{M}$

---

Step1: Find molar mass of $\text{AgNO}_3$

Molar mass of $\text{AgNO}_3 = 107.87 + 14 + 3\times16 = 169.87\ \text{g/mol}$

Step2: Calculate moles of $\text{AgNO}_3$

$n = \frac{10.0\ \text{g}}{169.87\ \text{g/mol}} \approx 0.0589\ \text{mol}$

Step3: Convert volume to liters

$V = 500\ \text{mL} = 0.500\ \text{L}$

Step4: Calculate molarity

$M = \frac{0.0589\ \text{mol}}{0.500\ \text{L}} \approx 0.118\ \text{M}$

---

Step1: Calculate moles of $\text{KNO}_3$

$n = M \times V = 0.500\ \text{M} \times 2.00\ \text{L} = 1.00\ \text{mol}$

Step2: Find molar mass of $\text{KNO}_3$

Molar mass of $\text{KNO}_3 = 39.1 + 14 + 3\times16 = 101.1\ \text{g/mol}$

Step3: Calculate mass of $\text{KNO}_3$

$m = n \times M = 1.00\ \text{mol} \times 101.1\ \text{g/mol} = 101.1\ \text{g}$

---

Step1: Find molar mass of KCl

Molar mass of $\text{KCl} = 39.1 + 35.5 = 74.6\ \text{g/mol}$

Step2: Calculate moles of KCl

$n = \frac{5.0\ \text{g}}{74.6\ \text{g/mol}} \approx 0.0670\ \text{mol}$

Step3: Calculate required volume

$V = \frac{n}{M} = \frac{0.0670\ \text{mol}}{0.25\ \text{M}} = 0.268\ \text{L} = 270\ \text{mL}$

---

Step1: Find molar mass of $\text{CuSO}_4\cdot5\text{H}_2\text{O}$

Molar mass $= 63.55 + 32.07 + 4\times16 + 5\times(2\times1 + 16) = 249.62\ \text{g/mol}$

Step2: Convert volume to liters

$V = 100\ \text{mL} = 0.100\ \text{L}$

Step3: Calculate moles of solute

$n = M \times V = 0.10\ \text{M} \times 0.100\ \text{L} = 0.010\ \text{mol}$

Step4: Calculate mass of hydrate

$m = 0.010\ \text{mol} \times 249.62\ \text{g/mol} = 2.50\ \text{g}$

---

Step1: Find molar mass of KI

Molar mass of $\text{KI} = 39.1 + 126.9 = 166\ \text{g/mol}$

Step2: Calculate moles of KI

$n = \frac{0.65\ \text{g}}{166\ \text{g/mol}} \approx 0.003916\ \text{mol}$

Step3: Calculate molarity

$M = \frac{0.003916\ \text{mol}}{0.300\ \text{L}} \approx 0.013\ \text{M}$

Answer:

1. $\approx 1.0\ \text{M}$
2. $\approx 0.118\ \text{M}$
3. $101\ \text{g}$ (rounded)
4. $\approx 0.27\ \text{L}$ (or $270\ \text{mL}$)
5. $2.5\ \text{g}$ (rounded)
6. $\approx 0.013\ \text{M}$