Assume 100 g of solution:
- Glucose = 10 g
- Water = 90 g = 0.090 kg
Molality:
\[
\text{Mol. mass of glucose} = 180 \text{ g/mol},\quad
\text{mol} = \frac{10}{180} = 0.0556
\]
\[
m = \frac{0.0556}{0.090} \approx 0.617 \approx 0.611 \text{ mol/kg}
\]
Molarity:
\[
\text{Volume} = \frac{100}{1.2} = 83.33 \text{ mL} = 0.08333 \text{ L}
\]
\[
M = \frac{0.0556}{0.08333} \approx 0.67 \text{ mol/L}
\]