Question

The molality and molarity of a solution of glucose in water which is labeled as \(10\%\) (w/w) are respectively \((\text{density of solution}=1.2\,\text{g mL}^{-1})\):

• A. \(0.57\,m,\;0.517\,M\)
• B. \(0.67\,m,\;0.617\,M\)
• C. \(0.617\,m,\;0.67\,M\)
• D. \(0.517\,m,\;0.57\,M\)

Hint:

For \(x\%\) (w/w) solutions, always assume \(100\,g\) of solution first. Then: \[ \text{Molality}=\frac{\text{moles of solute}}{\text{kg of solvent}} \] and \[ \text{Molarity}=\frac{\text{moles of solute}}{\text{volume of solution in litres}} \] Use density to convert mass of solution into volume.

Answer 1

The Correct Option is C

Step 1: Interpret the \(10\%\) (w/w) concentration.
A \(10\%\) (w/w) glucose solution means \[ 10\,\text{g} \] of glucose is present in \[ 100\,\text{g} \] of solution.
Therefore, mass of water is \[ 100-10=90\,\text{g} \] \[ =0.090\,\text{kg} \]

Step 2: Calculate the number of moles of glucose.
Molar mass of glucose \((C_6H_{12}O_6)\) is \[ 6(12)+12(1)+6(16) \] \[ =72+12+96 \] \[ =180\,\text{g mol}^{-1} \] Hence, \[ \text{Moles of glucose} = \frac{10}{180} \] \[ =0.0556\,\text{mol} \]

Step 3: Calculate molality.
Molality is given by \[ m=\frac{\text{moles of solute}} {\text{kg of solvent}} \] Substituting the values, \[ m=\frac{0.0556}{0.090} \] \[ m=0.617 \] Therefore, \[ \boxed{m=0.617\,m} \]

Step 4: Calculate molarity.
Density of solution is \[ 1.2\,\text{g mL}^{-1} \] Volume of \(100\,\text{g}\) solution is \[ V=\frac{\text{mass}}{\text{density}} \] \[ V=\frac{100}{1.2} \] \[ =83.33\,\text{mL} \] \[ =0.08333\,\text{L} \] Molarity is \[ M=\frac{\text{moles of solute}} {\text{volume of solution in litres}} \] Thus, \[ M=\frac{0.0556}{0.08333} \] \[ M=0.667 \] \[ M\approx 0.67 \] Therefore, \[ \boxed{M=0.67\,M} \]

Step 5: Final conclusion.
Hence, the molality and molarity respectively are \[ \boxed{0.617\,m,\;0.67\,M} \] Therefore, the correct option is \[ \boxed{(3)} \]