Question

1. a solution of cuso₄ is made by dissolving.15 g in enough water to make 150 ml of solution. what is the molarity of the solution?
2. 500 ml of solution is made from 3.12 g of ca(no₃)₂. what is the molarity of the solution?
3. how many grams of nacl would be needed to make 2.00 l of a 0.100 m solution?
4. hcl is purchased as a 12 m solution. how many ml of this stock solution would be needed to make 200 ml of.3 m solution?
5. a student made 50 ml of a.100 m solution using 13 ml of an unknown solution. what was the molarity of the unknown solution?

Explanation:

Question 14:

Step 1: Calculate moles of $\ce{CuSO4}$

Molar mass of $\ce{CuSO4}$ is $63.55 + 32.07 + 4\times16.00 = 159.62\ \text{g/mol}$.
Moles = $\frac{\text{mass}}{\text{molar mass}} = \frac{0.15\ \text{g}}{159.62\ \text{g/mol}} \approx 9.397\times10^{-4}\ \text{mol}$.

Step 2: Convert volume to liters

Volume = $150\ \text{mL} = 0.150\ \text{L}$.

Step 3: Calculate molarity

Molarity ($M$) = $\frac{\text{moles}}{\text{volume (L)}} = \frac{9.397\times10^{-4}\ \text{mol}}{0.150\ \text{L}} \approx 0.00626\ \text{M}$.

Step 1: Molar mass of $\ce{Ca(NO3)2}$

Molar mass = $40.08 + 2\times(14.01 + 3\times16.00) = 164.10\ \text{g/mol}$.

Step 2: Moles of $\ce{Ca(NO3)2}$

Moles = $\frac{3.12\ \text{g}}{164.10\ \text{g/mol}} \approx 0.01901\ \text{mol}$.

Step 3: Volume in liters

Volume = $500\ \text{mL} = 0.500\ \text{L}$.

Step 4: Molarity

$M = \frac{0.01901\ \text{mol}}{0.500\ \text{L}} \approx 0.0380\ \text{M}$.

Step 1: Moles of $\ce{NaCl}$ needed

Molarity $M = 0.100\ \text{M}$, Volume $V = 2.00\ \text{L}$.
Moles $n = M \times V = 0.100\ \text{mol/L} \times 2.00\ \text{L} = 0.200\ \text{mol}$.

Step 2: Molar mass of $\ce{NaCl}$

Molar mass = $22.99 + 35.45 = 58.44\ \text{g/mol}$.

Step 3: Mass of $\ce{NaCl}$

Mass = $n \times$ molar mass = $0.200\ \text{mol} \times 58.44\ \text{g/mol} = 11.688\ \text{g} \approx 11.7\ \text{g}$.

Answer:

$\approx 0.0063\ \text{M}$ (or $6.3\times10^{-3}\ \text{M}$)

Question 15: