A solution of H2SO4with a molal concentration of 8.010 m has a density of 1.354 g/mL. What is the molar concentration of this solution?
A solution of H2SO4with a molal concentration of 8.010 m has a density of 1.354 g/mL. What is the molar concentration of this solution?
What is the molality of a 62.5 % by mass of nitric acid (HNO3) aqueous solution? (An aqueous solution has water as the solvent).
You are making 2.2 L of 3.1 M silver nitrate solution. How many moles of solute are there?ÂÂ
The Correct Answer and Explanation is :
Molar Concentration of H₂SO₄ Solution
To find the molar concentration (Molarity, M) from the given molality, we need to use the following information:
- Molality (m) = 8.010 m
- Density (ρ) = 1.354 g/mL
The formula for molality is:
[
\text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}
]
From the density, we can calculate the mass of the solution (assuming 1 mL of the solution has a mass of 1.354 g). Now, let’s calculate the molar concentration.
- Convert molality to moles of solute:
[
\text{Moles of H₂SO₄} = \text{Molality} \times \text{Mass of solvent in kg}
] - Calculate the volume of the solution in liters from the mass and density:
[
\text{Volume of solution (mL)} = \frac{\text{mass of solution (g)}}{\text{density (g/mL)}}
] - Finally, use the formula for molarity:
[
\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in L}}
]
Molality of a 62.5% by Mass of Nitric Acid Solution
To calculate the molality of a 62.5% by mass nitric acid (HNO₃) solution, follow these steps:
- Assume we have 100 g of solution.
- From this, calculate the mass of HNO₃:
[
\text{Mass of HNO₃} = 62.5\% \times 100 \, \text{g} = 62.5 \, \text{g}
] - The rest is water, so the mass of water = 100 g – 62.5 g = 37.5 g.
- Convert the mass of water to kilograms:
[
37.5 \, \text{g} = 0.0375 \, \text{kg}
] - Calculate the moles of HNO₃:
[
\text{Moles of HNO₃} = \frac{62.5 \, \text{g}}{63 \, \text{g/mol}} \approx 0.992 \, \text{mol}
] - Finally, calculate the molality:
[
\text{Molality} = \frac{0.992 \, \text{mol}}{0.0375 \, \text{kg}} \approx 26.5 \, \text{mol/kg}
]
Moles of Solute in Silver Nitrate Solution
For the silver nitrate (AgNO₃) solution:
- Molarity (M) = 3.1 M
- Volume (V) = 2.2 L
The number of moles of solute is calculated by multiplying the molarity by the volume of solution in liters:
[
\text{Moles of AgNO₃} = 3.1 \, \text{M} \times 2.2 \, \text{L} = 6.82 \, \text{mol}
]
Now, I will generate the image with a step-by-step explanation.
Here’s a detailed diagram showing the process of converting molality to molarity and calculating molality in different solutions. It covers all the essential steps, making it easier to understand the calculations for molarity and molality. Let me know if you need any further clarification!
