Question

At \(T(K)\), \(x\) g of a non-volatile solid (molar mass \(78\ \mathrm{g\ mol^{-1}}\)) when added to \(0.5\) kg water, lowered its freezing point by \(1.0^\circ C\). What is \(x\) (in g)?
\[ K_f \text{ of water at } T(K)=1.86\ \mathrm{K\ kg\ mol^{-1}} \]

• A. \(10.48\)
• B. \(20.96\)
• C. \(41.92\)
• D. \(5.24\)

Hint:

For freezing point depression problems, always use \(\Delta T_f=K_f m\), where molality is moles of solute per kilogram of solvent.

Answer 1

The Correct Option is B

Step 1: Use depression in freezing point formula.
The depression in freezing point is given by
\[ \Delta T_f=K_f m \]
where \(m\) is molality.
Given,
\[ \Delta T_f=1.0\ \mathrm{K} \] \[ K_f=1.86\ \mathrm{K\ kg\ mol^{-1}} \]
Therefore,
\[ m=\frac{\Delta T_f}{K_f} \]
\[ m=\frac{1.0}{1.86} \]
\[ m\approx0.5376\ \mathrm{mol\ kg^{-1}} \]

Step 2: Calculate number of moles of solute.
Molality is defined as
\[ m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Mass of water \(=0.5\ \mathrm{kg}\).
Hence, moles of solute are
\[ \text{moles}=0.5376\times0.5 \]
\[ =0.2688\ \mathrm{mol} \]

Step 3: Calculate mass of solute.
Given molar mass of solute \(=78\ \mathrm{g\ mol^{-1}}\).
\[ x=\text{moles}\times\text{molar mass} \]
\[ x=0.2688\times78 \]
\[ x\approx20.96\ \mathrm{g} \]

Step 4: Final conclusion.
Hence, the value of \(x\) is
\[ \boxed{20.96\ \mathrm{g}} \]