Question

250 mL of a sucrose solution contains $1.51\times {10}^{23}$ molecules of the carbohydrate. The concentration of the solution is:

• A. 0.1 M
• B. 0.01 M
• C. 1 M
• D. 1.5 M

Answer 1

The correct option is C 1 M

1 mole of sucrose contains $N_A$ molecules.
$6.023\times {10}^{23}$ molecules are present in 1 mol of sucrose.
$1.51\times {10}^{23}\text{molecules will be in}\frac{1.51\times {10}^{23}}{6.023\times {10}^{23}}=0.25\text{mol}$

$\text{Molarity}=\frac{\text{number of moles of solute}}{\text{volume of solution in mL}}\times 1000$

$\text{Molarity}=\frac{0.25}{250}\times 1000=1M$