Q: Typical seawater contains 2.7 g of salt (sodium chloride, NaCl) per 100 mL. What is the molarity of NaCl in the ocean?

Here are the step-by-step solutions to the questions.

Q1.
Q: Typical seawater contains $2.7 \text{ g}$ of salt (sodium chloride, NaCl) per $100 \text{ mL}$. What is the molarity of NaCl in the ocean?

Step 1: Calculate the molar mass of NaCl.
Molar mass of Na = $22.99 \text{ g/mol}$
Molar mass of Cl = $35.45 \text{ g/mol}$
Molar mass of NaCl = $22.99 \text{ g/mol} + 35.45 \text{ g/mol} = 58.44 \text{ g/mol}$

Step 2: Calculate the moles of NaCl.
Given mass of NaCl = $2.7 \text{ g}$

$$\text{Moles of NaCl} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2.7 \text{ g}}{58.44 \text{ g/mol}} = 0.04619 \text{ mol}$$

Step 3: Convert the volume of seawater to liters.
Given volume = $100 \text{ mL}$

$$100 \text{ mL} = 100 \times 10^{-3} \text{ L} = 0.100 \text{ L}$$

Step 4: Calculate the molarity of NaCl.

$$\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}} = \frac{0.04619 \text{ mol}}{0.100 \text{ L}} = \boxed{\text{0.46 M}}$$

Q2.
Q: MgCl₂ has a concentration of $0.054 \text{ M}$ in the ocean. How many grams of MgCl₂ are present in $25 \text{ mL}$ of seawater?

Step 1: Convert the volume of seawater to liters.
Given volume = $25 \text{ mL}$

$$25 \text{ mL} = 25 \times 10^{-3} \text{ L} = 0.025 \text{ L}$$

Step 2: Calculate the moles of MgCl₂.
Given molarity = $0.054 \text{ M}$

$$\text{Moles of MgCl}_2 = \text{Molarity} \times \text{Volume (L)} = 0.054 \text{ mol/L} \times 0.025 \text{ L} = 0.00135 \text{ mol}$$

Step 3: Calculate the molar mass of MgCl₂.
Molar mass of Mg = $24.31 \text{ g/mol}$
Molar mass of Cl = $35.45 \text{ g/mol}$
Molar mass of MgCl₂ = $24.31 \text{ g/mol} + (2 \times 35.45 \text{ g/mol}) = 24.31 \text{ g/mol} + 70.90 \text{ g/mol} = 95.21 \text{ g/mol}$$

Step 4: Calculate the mass of MgCl₂.

$$\text{Mass of MgCl}_2 = \text{Moles} \times \text{Molar mass} = 0.00135 \text{ mol} \times 95.21 \text{ g/mol} = \boxed{\text{0.13 g}}$$

Q3.
Q: Find the molarity and molality of 37.0 \text{ wt% HCl}. Assume the density of the acid to be $1.19 \text{ g/mL}$.

Molarity Calculation:

Step 1: Assume a basis of $100 \text{ g}$ of solution.
Mass of HCl = $37.0 \text{ g}$ (since it's 37.0 \text{ wt%})
Mass of solution = $100 \text{ g}$

Step 2: Calculate the molar mass of HCl.
Molar mass of H = $1.008 \text{ g/mol}$
Molar mass of Cl = $35.45 \text{ g/mol}$
Molar mass of HCl = $1.008 \text{ g/mol} + 35.45 \text{ g/mol} = 36.458 \text{ g/mol}$

Step 3: Calculate the moles of HCl.

$$\text{Moles of HCl} = \frac{\text{Mass of HCl}}{\text{Molar mass of HCl}} = \frac{37.0 \text{ g}}{36.458 \text{ g/mol}} = 1.0149 \text{ mol}$$

Step 4: Calculate the volume of the solution.
Given density of solution = $1.19 \text{ g/mL}$

$$\text{Volume of solution} = \frac{\text{Mass of solution}}{\text{Density of solution}} = \frac{100 \text{ g}}{1.19 \text{ g/mL}} = 84.03 \text{ mL}$$

Step 5: Convert the volume to liters.

$$84.03 \text{ mL} = 84.03 \times 10^{-3} \text{ L} = 0.08403 \text{ L}$$

Step 6: Calculate the molarity.

$$\text{Molarity (M)} = \frac{\text{Moles of HCl}}{\text{Volume of solution (L)}} = \frac{1.0149 \text{ mol}}{0.08403 \text{ L}} = \boxed{\text{12.1 M}}$$

Molality Calculation:

Step 1: Determine the mass of the solvent (water).
Mass of solution = $100 \text{ g}$
Mass of HCl (solute) = $37.0 \text{ g}$
Mass of solvent = Mass of solution - Mass of solute = $100 \text