An aqueous solution of glucose is labelled as 20% w/w . If the densit…

The correct answer is option 4. \(1.38\, \ mol/kg\).

To find the molality (\(m\)), we can use the following formula:
\(m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}\)

Given that the solution is \(20\% \ w/w\), it means that \(20\%\) of the total solution mass is glucose, and the rest is water.

Let's assume we have a \(100 \ g\) solution. In that case, \(20\%\) of \(100 \ g\) is glucose, and \(80\%\) is water.

\(\text{Mass of glucose} = 20\% \times 100 \ g = 20 \ g\)

\(\text{Mass of water} = 80\% \times 100 \ g = 80 \ g\)

Now, we can calculate the moles of solute (glucose) using its molar mass, and then find the molality.

\(n = \frac{\text{Mass of solute}}{\text{Molar mass of glucose}}\)

The molar mass of glucose (\(C_6H_{12}O_6\)) is approximately \(180 \ g/mol\).

\(n = \frac{20 \ g}{180 \ g/mol}\)

\(n \approx 0.1111 \ mol\)  

\(m = \frac{n}{\text{Mass of solvent (in kg)}}\)

Since the density of the solution is \(1.2 \ g/mL\), the mass of \(80 \ g\) of water is:

\(\text{Mass of water (in kg)} = \frac{80 \ g}{1000 \ g/kg} = 0.08 \ kg\)

\(m = \frac{0.1111 \ mol}{0.08 \ kg} \approx 1.38 \ mol/kg\)

So the correct answer is: 4. \(1.38 \ mol/kg\)