Question

question 18 of 30
what is the change in the freezing point of water when 35.0 g of sucrose is
dissolved in 300.0 g of water?

$k_f$ of water = -1.86°c/mol
molar mass sucrose = 342.30 g/mol
$i$ value of sugar = 1

\\( \bigcirc \\) a. -0.634°c
\\( \bigcirc \\) b. 0.0606 °c
\\( \bigcirc \\) c. -0.183°c
\\( \bigcirc \\) d. 0.217°c

Explanation:

Step1: Calculate moles of sucrose

Moles = $\frac{\text{mass}}{\text{molar mass}}$ = $\frac{35.0\ \text{g}}{342.30\ \text{g/mol}}$ ≈ 0.10225 mol

Step2: Calculate molality (m)

Molality = $\frac{\text{moles of solute}}{\text{kg of solvent}}$ = $\frac{0.10225\ \text{mol}}{0.3000\ \text{kg}}$ ≈ 0.3408 mol/kg

Step3: Calculate freezing point change ($\Delta T_f$)

$\Delta T_f = i \times K_f \times m$ = $1 \times (-1.86\ ^\circ\text{C/mol}) \times 0.3408\ \text{mol/kg}$ ≈ -0.634 $^\circ$C

Answer:

A. -0.634$^\circ$C