Question

If \(50\%\) of \(1\ M\ Na_2SO_4\) is dissociated in aqueous solution of density \(1.2\ g\ mL^{-1}\), what is the molality of \(Na^+\) ions in the solution?

• A. \(0.95\)
• B. \(1.89\)
• C. \(1.00\)
• D. \(2.00\)

Hint:

To convert molarity into molality, first calculate the mass of solvent using density of the solution.

Answer 1

The Correct Option is A

Step 1: Understand the given data.
Molarity of solution: \[ 1\ M \] Thus, \[ 1\ \text{L solution contains }1\ \text{mol of }Na_2SO_4 \] Density of solution: \[ 1.2\ g\ mL^{-1} \] Therefore, mass of \(1\ L\) solution is \[ 1000\times1.2=1200\ g \]

Step 2: Calculate mass of solvent.
Molar mass of \(Na_2SO_4\): \[ (2\times23)+32+(4\times16)=142\ g\ mol^{-1} \] Mass of solute present: \[ 142\ g \] Hence, mass of water: \[ 1200-142=1058\ g \] \[ =1.058\ kg \]

Step 3: Calculate moles of \(Na^+\) ions formed.
Dissociation: \[ Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} \] Degree of dissociation: \[ \alpha=50\%=0.5 \] From \(1\ mol\) of \(Na_2SO_4\), \[ 2\times0.5=1\ mol \] of \(Na^+\) ions are formed.

Step 4: Calculate molality.
Molality is \[ m=\frac{\text{moles of solute}}{\text{kg of solvent}} \] Thus, \[ m=\frac{1}{1.058} \approx0.95 \]

Step 5: Final conclusion.
Hence, molality of \(Na^+\) ions is \[ \boxed{0.95} \]