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: Find moles of glucose and mass of solvent.
10% (w/w): 10 g glucose in 100 g solution. Solvent = 90 g = 0.09 kg. Moles of glucose = 10/180 = 0.0556 mol.

Step 2: Calculate molality then molarity.
Molality = 0.0556/0.09 = 0.617 m. Volume of solution = 100 g / 1.2 g/mL = 83.33 mL = 0.08333 L. Molarity = 0.0556/0.08333 = 0.667 M \(\approx\) 0.67 M. \[ oxed{0.617\,m,\ 0.67\,M} \]