Practicing Success
An aqueous solution of glucose is labelled as \(20\%\) \(w/w\). If the density of the solution is \(1.2\, \ g/mL\) the molality of the solution will be |
\(5.38\, \ mol/kg\) \(4.21\, \ mol/kg\) \(2.76\, \ mol/kg\) \(1.38\, \ mol/kg\) |
\(1.38\, \ mol/kg\) |
The correct answer is option 4. \(1.38\, \ mol/kg\). To find the molality (\(m\)), we can use the following formula: 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. \(1.38 \ mol/kg\) |