Molarity (M) of an aqueous solution containing \(x \, \text{g}\) of anhyd. CuSO$_4$ in 500 mL solution at 32$^\circ$C is \(2 \times 10^{-1} \, \text{M}\). Its molality will be ____ \( \times 10^{-3} \, \text{m}\) (nearest integer).
[Given density of the solution = 1.25 g/mL.]
[Given density of the solution = 1.25 g/mL.]
Correct Answer: 164
Approach Solution - 1
To find the molality of the solution, we first need to determine the number of moles of anhydrous CuSO4 in the solution. Given the molarity (M) is \(2 \times 10^{-1} \, \text{M}\) in a 500 mL solution, we calculate the moles of CuSO4 as follows: \( \text{Moles of CuSO}_4 = \text{Molarity} \times \text{Volume in L} = 2 \times 10^{-1} \times 0.5 = 0.1 \, \text{mol} \).
The molecular weight of CuSO4 is \(63.5 + 32 + 4 \times 16 = 159.5 \, \text{g/mol}\). Therefore, \( x = \text{moles} \times \text{molecular weight} = 0.1 \times 159.5 = 15.95 \, \text{g}\).
Next, calculate the mass of the solution using the given density: \( \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.25 \times 500 = 625 \, \text{g} \).
Subtracting the mass of CuSO4, the mass of the solvent (water) is: \(625 - 15.95 = 609.05 \, \text{g} \) or 0.60905 kg.
The molality (m) is defined as moles of solute per kg of solvent: \( \text{Molality} = \frac{\text{Moles of CuSO}_4}{\text{kg of solvent}} = \frac{0.1}{0.60905} \approx 0.164 \, \text{mol/kg} \) or \(164 \times 10^{-3} \, \text{m}\).
Thus, the molality of the solution is \(164 \times 10^{-3} \, \text{m}\), which is within the expected range of 164,164.
Approach Solution -2
Given:
- Molarity (\(M\)) = 0.2 mol/L
- Volume of solution (\(V_{sol}\)) = 500 mL = 0.5 L
- Density of solution (\(d_{sol}\)) = 1.25 g/mL
- Molar mass of CuSO\(_4\) = 159.5 g/mol
Step 1: Calculate the mass of the solution
\[ M_{sol} = V_{sol} \times d_{sol} = 500 \, \text{mL} \times 1.25 \, \text{g/mL} = 625 \, \text{g}. \]
Step 2: Calculate the mass of solute
\[ \text{Mass of solute (CuSO}_4\text{)} = M \times V_{sol} \times \text{Molar mass}. \] \[ = 0.2 \times 0.5 \times 159.5 = 15.95 \, \text{g}. \]
Step 3: Calculate the mass of the solvent
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute}. \] \[ = 625 - 15.95 = 609.05 \, \text{g} = 0.60905 \, \text{kg}. \]
Step 4: Calculate the molality
\[ m = \frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}} = \frac{0.1}{0.60905}. \] \[ m = 0.164 \, \text{mol/kg} = 164 \times 10^{-3} \, \text{mol/kg}. \]
Final Answer
The molality of the solution is:
\[ 0.164 \, \text{mol/kg (or } 164 \times 10^{-3} \, \text{mol/kg)}. \]
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