Question

What is the concentration of sugar$\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)$ in ${\text{mol L}}^{-1}$ if its $20\text{g}$ are dissolved in enough water to make up a final volume up to $2\text{L}$?

Answer 1

Step1. Molarity

• "Molarity ($M$) is defined as the number of moles of solute dissolved in one liter (or one cubic decimetre) of the solution."

Step2. Formula for molarity

$\text{Molarity =}\frac{\text{Number of moles of solute}}{\mathrm{Volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{liters}}$

Step3.Calculation of molar mass of sugar

The molar mass of sugar=

$$\begin{gathered} =12\times \text{Atomic mass of C}+22\times \text{Atomic mass of H+11×}\mathrm{Atomic}\mathrm{mass}\mathrm{of}\text{O} \\ =\left(12\times 12+22\times 1+16\times 11\right){\text{g mol}}^{-1} \\ =\left(144+22+176\right){\text{g mol}}^{-1} \\ =342{\text{g mol}}^{-1} \end{gathered}$$

Step4. Calculation of number of moles

$$\begin{gathered} \text{Moles=}\frac{\text{Mass}}{\text{Molar mass}} \\ \text{=}\frac{\text{20 g}}{342{\text{g mol}}^{-1}} \\ \text{=0.0584 mol} \end{gathered}$$

Step5. Calculation of molarity

$$\begin{gathered} \text{Molarity =}\frac{\text{0.0584 mol}}{2\text{L}} \\ {\text{=0.029 mol L}}^{\text{-1}} \\ \text{≃0.03}\mathrm{mol}{\text{L}}^{\text{-1}} \end{gathered}$$

Hence, the concentration of sugar is ${\text{0.03 mol L}}^{\text{-1}}$.