Kimyaa soru 12

YKS TYT
1

IMG_20250104_093300
IMG_20250104_093300719×216 21.4 KB

2

What is the molality of a solution prepared by dissolving 171 grams of sucrose (C₁₂H₂₂O₁₁) in 200 grams of water?

Answer:

Molality ((m)) is defined as the number of moles of solute per kilogram of solvent. The formula to calculate molality is:

m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}

Step 1: Find the Molar Mass of Sucrose (C₁₂H₂₂O₁₁)

To find the molar mass of sucrose, we’ll sum up the atomic masses of all the atoms in the formula:

  • Carbon (C): 12 atoms × 12 g/mol = 144 g/mol
  • Hydrogen (H): 22 atoms × 1 g/mol = 22 g/mol
  • Oxygen (O): 11 atoms × 16 g/mol = 176 g/mol

Total molar mass of C₁₂H₂₂O₁₁ = 144 + 22 + 176 = 342 g/mol

Step 2: Calculate the Moles of Sucrose

Given mass of sucrose = 171 g

Using the molar mass from Step 1, calculate the moles of sucrose:

\text{moles of sucrose} = \frac{\text{mass of sucrose}}{\text{molar mass of sucrose}} = \frac{171 \text{ g}}{342 \text{ g/mol}} = 0.5 \text{ moles}

Step 3: Convert the Mass of Water to Kilograms

Given mass of water = 200 g

Convert to kilograms:

\text{mass of water in kg} = \frac{200 \text{ g}}{1000 \text{ g/kg}} = 0.2 \text{ kg}

Step 4: Calculate the Molality

Using the formula for molality:

m = \frac{\text{moles of sucrose}}{\text{kilograms of water}} = \frac{0.5 \text{ moles}}{0.2 \text{ kg}} = 2.5 \text{ mol/kg}

The calculation for molality quoted in the original list seems incorrect, as our computed molality is 2.5 mol/kg, which doesn’t match the multiple-choice options. Verify that all values used are accurate, or double-check the options provided. In its technical precision, the determined molality indicates further steps are correct, given no typos mount from the original image.

Final Answer:

The molality of the solution is 2.5 mol/kg. It looks like the possible options shown should be re-examined based on proper calculation or, perhaps, there may be discrepancies within the original choices provided.