Question

$500\text{ mL}$ of an aqueous solution of glucose $\text{C}_6\text{H}_{12}\text{O}_6$ (Molar mass $180\text{ g mol}^{-1}$) contains $02 \times 10^{22}$ molecules. The concentration of the solution will be:

• A. $2.0\text{ M}$
• B. $1.0\text{ M}$
• C. $0.2\text{ M}$
• D. $0.1\text{ M}$

Hint:

Notice the exponents in the scientific notation: $10^{22}$ divided by $10^{23}$ is exactly $0.1\text{ moles}$. Since $0.1\text{ moles}$ are dissolved in half a liter ($500\text{ mL}$), doubling it to find the amount per full liter gives a concentration of $0.2\text{ M}$ in seconds.

Answer 1

The Correct Option is C

Concept: Molarity ($M$) measures the concentration of a solute in a solution, defined as the number of moles of solute ($n$) dissolved per liter of total solution volume ($V$): \[ M = \frac{n}{V\text{ (in Liters)}} \] The number of moles can be calculated from the total number of particles using Avogadro's number ($N_{\text{A}} = 6.023 \times 10^{23}\text{ molecules mol}^{-1}$).

Step 1: Calculate the number of moles of glucose solute ($n$).
Given particle count $= 6.02 \times 10^{22}$ molecules: \[ n = \frac{\text{Number of molecules}}{N_{\text{A}}} = \frac{6.02 \times 10^{22}}{6.023 \times 10^{23}} \approx 0.1\text{ moles} \]

Step 2: Convert the solution volume to liters and calculate molarity.
Given volume $V = 500\text{ mL} = 0.5\text{ L}$: \[ M = \frac{0.1\text{ moles}}{0.5\text{ Liters}} = \frac{1}{5} = 0.2\text{ mol L}^{-1} = 0.2\text{ M} \]