1- Calculate the mass of NaOH which is required to prepare 0.1 M in 250 cm3 aqueous solution (molar mass = 40 gm/mole) 2- Calculate the mass of HCl which is required to prepare 0.1 N in 500 ml aqueous solution. (molar mass = 36.5 gm/mole) 3- 10 gm KCl dissolved in 400 cm3 aqueous solution. Calculate the molarity and molality of the solution. (molar mass, KCl = 74.5 gm/mole – density of the solution = 1.1 gm/ cm3 ). PROBLEMS 4) Calculate the molality of a sulfuric acid solution containing 24.4 g of sulfuric acid in 198 g of water. The molar mass of sulfuric acid is 98.09 g. 5) The density of a 2.45 M aqueous solution of methanol (CH3OH) is 0.976 g/mL. What is the molality of the solution? The molar mass of methanol is 32.04 g. 6) Calculate the molality of a 35.4 percent (by mass) aqueous solution of phosphoric acid (H3PO4 ). The molar mass of phosphoric acid is 97.99 g 7) A solution containing 53 g of sodium carbonate (mol mass = 106) in 1 Kg water at 20°C. The density of the solution at this temperature is 1.1 g/ml. Determine for the solution. 1) the mole fraction of each component 1) the mass percent 2) the molarity 3) the molality 4) the normality.
The Correct Answer and Explanation is :
1) Mass of NaOH required to prepare 0.1 M in 250 cm³ aqueous solution (Molar mass = 40 g/mol)
Molarity (M) is defined as moles of solute per liter of solution. To calculate the mass of NaOH, use the formula:
[
\text{Moles of NaOH} = M \times V
]
Where:
- ( M = 0.1 \, \text{M} )
- ( V = 250 \, \text{cm}^3 = 0.25 \, \text{L} )
[
\text{Moles of NaOH} = 0.1 \, \text{mol/L} \times 0.25 \, \text{L} = 0.025 \, \text{mol}
]
Now, to find the mass:
[
\text{Mass of NaOH} = \text{Moles of NaOH} \times \text{Molar mass of NaOH}
]
[
\text{Mass of NaOH} = 0.025 \, \text{mol} \times 40 \, \text{g/mol} = 1 \, \text{g}
]
So, 1 gram of NaOH is required.
2) Mass of HCl required to prepare 0.1 N in 500 mL aqueous solution (Molar mass = 36.5 g/mol)
Normality (N) is defined as the number of equivalents per liter of solution. For HCl, 1 mole of HCl provides 1 equivalent, so normality is the same as molarity.
We need the mass of HCl to prepare 0.1 N in 500 mL (0.5 L) solution:
[
\text{Equivalents of HCl} = N \times V = 0.1 \, \text{eq/L} \times 0.5 \, \text{L} = 0.05 \, \text{eq}
]
Now, the mass:
[
\text{Mass of HCl} = \text{Equivalents of HCl} \times \text{Molar mass of HCl}
]
[
\text{Mass of HCl} = 0.05 \, \text{eq} \times 36.5 \, \text{g/mol} = 1.825 \, \text{g}
]
So, 1.825 grams of HCl is required.
3) Molarity and Molality of KCl solution
Given:
- Mass of KCl = 10 g
- Volume of solution = 400 cm³ = 0.4 L
- Molar mass of KCl = 74.5 g/mol
- Density = 1.1 g/cm³
Molarity (M):
[
\text{Moles of KCl} = \frac{\text{Mass of KCl}}{\text{Molar mass of KCl}} = \frac{10 \, \text{g}}{74.5 \, \text{g/mol}} = 0.134 \, \text{mol}
]
[
\text{Molarity} = \frac{\text{Moles of KCl}}{\text{Volume of solution (in L)}} = \frac{0.134 \, \text{mol}}{0.4 \, \text{L}} = 0.335 \, \text{M}
]
Molality (m):
Molality is moles of solute per kilogram of solvent. First, we need to find the mass of the solvent (water). The total mass of the solution is:
[
\text{Mass of solution} = \text{Density} \times \text{Volume of solution} = 1.1 \, \text{g/cm}^3 \times 400 \, \text{cm}^3 = 440 \, \text{g}
]
Mass of water (solvent):
[
\text{Mass of water} = 440 \, \text{g} – 10 \, \text{g} = 430 \, \text{g} = 0.430 \, \text{kg}
]
Now, calculate the molality:
[
\text{Molality} = \frac{\text{Moles of KCl}}{\text{Mass of solvent (in kg)}} = \frac{0.134 \, \text{mol}}{0.430 \, \text{kg}} = 0.312 \, \text{mol/kg}
]
4) Molality of sulfuric acid solution
Given:
- Mass of sulfuric acid = 24.4 g
- Mass of water = 198 g
- Molar mass of sulfuric acid = 98.09 g/mol
First, calculate the moles of sulfuric acid:
[
\text{Moles of H}_2\text{SO}_4 = \frac{24.4 \, \text{g}}{98.09 \, \text{g/mol}} = 0.249 \, \text{mol}
]
Now, calculate the molality:
[
\text{Molality} = \frac{0.249 \, \text{mol}}{0.198 \, \text{kg}} = 1.26 \, \text{mol/kg}
]
5) Molality of a 2.45 M aqueous solution of methanol
Given:
- Molarity = 2.45 M
- Density = 0.976 g/mL
- Molar mass of methanol = 32.04 g/mol
First, calculate the mass of 1 L of solution:
[
\text{Mass of solution} = \text{Density} \times \text{Volume} = 0.976 \, \text{g/mL} \times 1000 \, \text{mL} = 976 \, \text{g}
]
Now, calculate the moles of methanol in 1 L:
[
\text{Moles of methanol} = 2.45 \, \text{mol/L} \times 1 \, \text{L} = 2.45 \, \text{mol}
]
Mass of methanol:
[
\text{Mass of methanol} = 2.45 \, \text{mol} \times 32.04 \, \text{g/mol} = 78.5 \, \text{g}
]
Now, mass of water:
[
\text{Mass of water} = 976 \, \text{g} – 78.5 \, \text{g} = 897.5 \, \text{g} = 0.8975 \, \text{kg}
]
Finally, calculate the molality:
[
\text{Molality} = \frac{2.45 \, \text{mol}}{0.8975 \, \text{kg}} = 2.73 \, \text{mol/kg}
]
6) Molality of a 35.4% (by mass) aqueous solution of phosphoric acid
Given:
- Mass percent of H₃PO₄ = 35.4%
- Mass of solution = 100 g (for simplicity)
- Mass of H₃PO₄ = 35.4 g
- Molar mass of H₃PO₄ = 97.99 g/mol
Mass of water:
[
\text{Mass of water} = 100 \, \text{g} – 35.4 \, \text{g} = 64.6 \, \text{g} = 0.0646 \, \text{kg}
]
Moles of H₃PO₄:
[
\text{Moles of H₃PO₄} = \frac{35.4 \, \text{g}}{97.99 \, \text{g/mol}} = 0.361 \, \text{mol}
]
Molality:
[
\text{Molality} = \frac{0.361 \, \text{mol}}{0.0646 \, \text{kg}} = 5.59 \, \text{mol/kg}
]
7) Solution containing 53 g of sodium carbonate (Na₂CO₃) in 1 kg of water
Given:
- Mass of Na₂CO₃ = 53 g
- Mass of water = 1 kg
- Molar mass of Na₂CO₃ = 106 g/mol
- Density of solution = 1.1 g/mL
Mole fraction:
[
\text{Moles of Na₂CO₃} = \frac{53 \, \text{g}}{106 \, \text{g/mol}} = 0.5 \, \text{mol}
]
Total mass of the solution:
[
\text{Mass of solution} = 53 \, \text{g} + 1000 \, \text{g} = 1053 \, \text{g}
]
Moles of water:
[
\text{Moles of water} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} = 55.56 \, \text{mol}
]
Mole fraction of Na₂CO₃:
[
\chi_{\text{Na₂CO₃}} = \frac{0.5}{0.5 + 55.56} = 0.00896
]
Mole fraction of water:
[
\chi_{\text{H₂O}} = \frac{55.56}{0.5 + 55.56} = 0.991
]
Mass percent of Na₂CO₃:
[
\text{Mass percent of Na₂CO₃} = \frac{53 \, \text{g}}{1053 \, \text{g}} \times 100 = 5.03\%
]
Molarity:
[
\text{Molarity} = \frac{0.5 \, \text{mol}}{1 \, \text{L}} = 0.5 \, \text{M}
]
Molality:
[
\text{Molality} = \frac{0.5 \, \text{mol}}{1 \, \text{kg}} = 0.5 \, \text{mol/kg}
]
Normality:
[
\text{Normality} = 0.5 \, \text{N} \quad (\text{since Na₂CO₃ provides 2 equivalents of Na+ ions})
]