Calculate the molarity of a 30.0% hydrogen peroxide solution.The reported density of 30% hydrogen peroxide is 1.135 g/cm3.(H2O2= 34.0147 g/mol) a) How many grams of solution in 1L? b) howmany grams in hydrogen peroxide are in 1 L? c) What is the molarityand molality of the solution?
The Correct Answer and Explanation is :
To calculate the molarity and molality of a 30.0% hydrogen peroxide solution, we’ll go through a series of steps using the given data:
- The density of the solution is 1.135 g/cm³.
- The concentration of the solution is 30.0% by mass.
- The molar mass of hydrogen peroxide (H₂O₂) is 34.0147 g/mol.
Let’s calculate each part step by step.
a) How many grams of solution in 1 L?
We’re given the density of the solution as 1.135 g/cm³. Since 1 L = 1000 cm³, we can calculate the mass of 1 L of solution:
[
\text{Mass of 1 L of solution} = \text{Density} \times \text{Volume}
]
[
\text{Mass of 1 L of solution} = 1.135 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1135 \, \text{g}
]
Thus, 1 L of solution weighs 1135 grams.
b) How many grams of hydrogen peroxide are in 1 L?
Since the solution is 30.0% by mass hydrogen peroxide, the mass of H₂O₂ in 1 L of solution is:
[
\text{Mass of H₂O₂} = \frac{30.0}{100} \times 1135 \, \text{g} = 340.5 \, \text{g}
]
So, there are 340.5 grams of hydrogen peroxide in 1 L of solution.
c) Molarity and Molality of the solution
Molarity:
Molarity is defined as moles of solute per liter of solution. To calculate the molarity, we first need to convert the mass of hydrogen peroxide to moles:
[
\text{Moles of H₂O₂} = \frac{\text{Mass of H₂O₂}}{\text{Molar mass of H₂O₂}}
]
[
\text{Moles of H₂O₂} = \frac{340.5 \, \text{g}}{34.0147 \, \text{g/mol}} = 10.0 \, \text{mol}
]
Since the solution is 1 L, the molarity (M) is:
[
\text{Molarity} = \frac{10.0 \, \text{mol}}{1 \, \text{L}} = 10.0 \, \text{M}
]
Molality:
Molality is defined as moles of solute per kilogram of solvent. We need to find the mass of the solvent, which is the mass of water in the solution. The mass of water is:
[
\text{Mass of water} = \text{Mass of solution} – \text{Mass of H₂O₂} = 1135 \, \text{g} – 340.5 \, \text{g} = 794.5 \, \text{g} = 0.7945 \, \text{kg}
]
Now, we can calculate the molality (m):
[
\text{Molality} = \frac{10.0 \, \text{mol}}{0.7945 \, \text{kg}} = 12.6 \, \text{mol/kg}
]
Summary:
- The mass of the solution in 1 L is 1135 g.
- The mass of hydrogen peroxide in 1 L is 340.5 g.
- The molarity of the solution is 10.0 M.
- The molality of the solution is 12.6 mol/kg.