At what temperature a solution containing \(5.60 g\) of glucose per \(1000 g\) of water would boil at \(1.013\, \ bar\)? (k for water is \(0.52\, \ K\, \ kg\, \ mol^{-1}\)).

373.166 K

The correct answer is option 4. 373.166 K.

To calculate the boiling point elevation, we can use the formula:

\[ \Delta T_b = i \times K_b \times m \]

Where:

\( \Delta T_b \) is the boiling point elevation,

\( i \) is the van't Hoff factor,

\( K_b \) is the ebullioscopic constant (molal boiling point elevation constant),

\( m \) is the molality of the solute.

Given that the solution contains \(5.60 \, g\) of glucose per \(1000 \, g\) of water, we need to find the molality (\( m \)) of the solution.

The molar mass of glucose (\( C_6H_{12}O_6 \)) is \( 180.16 \, g/mol \).

\[ \text{Moles of glucose} = \frac{5.60 \, g}{180.16 \, g/mol} \approx 0.0311 \, mol \]

Since the mass of water is \(1000 \, g\), the mass of the solvent (water) is \(1000 - 5.60 = 994.40 \, g\).

\[ \text{Molality} \ (m) = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{0.0311 \, mol}{0.9944 \, kg} \approx 0.0313 \, mol/kg \]

Now, we can calculate the boiling point elevation (\( \Delta T_b \)):

\[ \Delta T_b = i \times K_b \times m \]

Given that water is the solvent, \( i = 1 \).

Given \( K_b = 0.52 \, K \, kg \, mol^{-1} \).

\[ \Delta T_b = 1 \times 0.52 \, K \, kg \, mol^{-1} \times 0.0313 \, mol/kg \]
\[ \Delta T_b \approx 0.0164 \, K \]

Finally, we can calculate the boiling point of the solution:

\[ T_{b \text{, solution}} = T_{b \text{, pure solvent}} + \Delta T_b \]

The boiling point of pure water at \(1.013 \, \text{bar}\) is \(373.15 \, K\).

\[ T_{b \text{, solution}} = 373.15 \, K + 0.0164 \, K \]

\[ T_{b \text{, solution}} \approx 373.166 \, K \]

So, the boiling point of the solution is approximately \(373.166 \, K\). Therefore, the correct answer is option 4.