Let’s solve each of the chemistry problems step by step.
1. Solubility of Carbon Dioxide in Water
We can use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the liquid. The formula is given as:
\[ S_2 = S_1 \times \left( \frac{P_2}{P_1} \right) \]
where:
- \( S_1 = 0.16 , \text{g/100 mL} \) (solubility at 1 atm)
- \( P_1 = 5.5 , \text{atm} \) (initial pressure)
- \( P_2 = 100 , \text{atm} \) (new pressure)
First, convert the solubility \( S_1 \) from grams per 100 mL to grams per liter (since there are 10 times more grams in 1000 mL):
\[ S_1 = 0.16 , \text{g/100 mL} \times 10 = 1.6 , \text{g/L} \]
Now, we can substitute into Henry's Law:
\[ S_2 = 1.6 , \text{g/L} \times \left( \frac{100 , \text{atm}}{5.5 , \text{atm}} \right) \]
\[ S_2 = 1.6 , \text{g/L} \times 18.18 \approx 29.09 , \text{g/L} \]
So, the solubility of carbon dioxide at 100 atm is approximately 29.09 g/L.
2. Molarity Calculation
A. \(10.5 , \text{g NaCl in 350.0 , mL}\)
Firstly, calculate the molar mass of NaCl:
\[ \text{Na (22.99 g/mol) + Cl (35.45 g/mol) = 58.44 g/mol} \]
Now find moles of NaCl:
\[ \text{Moles of NaCl} = \frac{10.5 , \text{g}}{58.44 , \text{g/mol}} \approx 0.179 , \text{mol} \]
Convert 350.0 mL to L:
\[ 350.0 , \text{mL} = 0.350 , \text{L} \]
Now calculate the molarity:
\[ \text{Molarity (M)} = \frac{\text{moles}}{\text{volume in L}} = \frac{0.179 , \text{mol}}{0.350 , \text{L}} \approx 0.511 , \text{M} \]
B. \(40.7, \text{g LiCl·3H2O in 1 L}\)
Calculate the molar mass of LiCl·3H2O:
- Li = 6.94 g/mol
- Cl = 35.45 g/mol
- 3 H2O = 3 × (2 + 16) = 54 g/mol
\[ \text{Molar Mass of LiCl·3H2O} = 6.94 + 35.45 + 54 = 96.39 , \text{g/mol} \]
Finding moles of LiCl·3H2O:
\[ \text{Moles} = \frac{40.7 , \text{g}}{96.39 , \text{g/mol}} \approx 0.422 , \text{mol} \]
Now calculate the molarity:
\[ \text{Molarity (M)} = \frac{0.422 , \text{mol}}{1 , \text{L}} \approx 0.422 , \text{M} \]
3. Percentage by Mass of H2SO4 in Solution
To find the percentage by mass, we need to first find the mass of H2SO4 within 1L of the solution:
- Given density = 1.338 g/mL, for 1 L (1000 mL):
\[ \text{Mass of solution} = 1.338 , \text{g/mL} \times 1000 , \text{mL} = 1338 , \text{g} \]
- Molarity of the solution = 6 mol/L, thus:
\[ \text{Moles of H2SO4} = 6 , \text{mol} \]
Molar mass of H2SO4:
\[ \text{H2SO4} = 2(1.01) + 32.07 + 4(16) = 98.09 , \text{g/mol} \]
Mass of H2SO4:
\[ \text{Mass of H2SO4} = 6 , \text{mol} \times 98.09 , \text{g/mol} = 588.54 , \text{g} \]
Now calculate the percentage by mass:
\[ \text{Percentage by mass} = \left( \frac{588.54 , \text{g}}{1338 , \text{g}} \right) \times 100 \approx 43.92% \]
4. Mass of Calcium Chloride for 0.125 mol/kg in 800g of Water
To find the mass of CaCl2 needed for a 0.125 mol/kg concentration in 800 g of water:
\[ 0.125 , \text{mol/kg} \times 0.800 , \text{kg} = 0.1 , \text{mol} \]
Molar mass of CaCl2 (Calcium = 40.08, Chlorine = 35.45):
\[ \text{CaCl2} = 40.08 + 2(35.45) = 111.98 , \text{g/mol} \]
Mass required:
\[ \text{Mass} = 0.100 , \text{mol} \times 111.98 , \text{g/mol} \approx 11.20 , \text{g} \]
5. Approximate Volume of Water to Dilute Solution
To dilute a solution:
- Use \( C_1V_1 = C_2V_2 \):
Let \( C_1 = 1.25N \), \( V_1 = 250 , \text{mL} \), \( C_2 = 0.5N \), and let \( V_2 \) be the final volume.
\[ 1.25N \times 250 , \text{mL} = 0.5N \times V_2 \]
Solving for \( V_2 \):
\[ 312.5N , \text{mL} = 0.5N \times V_2 \quad \Rightarrow \quad V_2 = \frac{312.5N}{0.5N} = 625 , \text{mL} \]
The additional volume of water to be added is:
\[ V_{\text{water}} = V_2 - V_1 = 625 , \text{mL} - 250 , \text{mL} = 375 , \text{mL} \]
Summary of Results
- Carbon Dioxide Solubility: ~29.09 g/L
- Molarity:
- A: ~0.511 M
- B: ~0.422 M
- Percentage by Mass of H2SO4: ~43.92%
- Mass of CaCl2 Required: ~11.20 g
- Volume of Water to Add: ~375 mL