Answer :
To solve the problem of determining the mass of urea dissolved in the substance [tex]\(X\)[/tex], let's follow these steps using the provided information:
1. Identify the Known Values:
- Normal freezing point of substance [tex]\(X\)[/tex]: [tex]\(-7.5^\circ \text{C}\)[/tex]
- New freezing point of the solution: [tex]\(-9.3^\circ \text{C}\)[/tex]
- Freezing point depression constant, [tex]\(K_f\)[/tex]: [tex]\(6.12 \, ^\circ \text{C} \cdot \text{kg/mol}\)[/tex]
- Mass of solvent, [tex]\(X\)[/tex]: [tex]\(100.8 \, \text{g} = 0.1008 \, \text{kg}\)[/tex]
2. Calculate the Freezing Point Depression:
[tex]\[
\Delta T_f = \text{Normal freezing point} - \text{New freezing point} = -7.5^\circ \text{C} - (-9.3^\circ \text{C}) = 1.8^\circ \text{C}
\][/tex]
3. Use the Formula for Freezing Point Depression:
[tex]\[
\Delta T_f = K_f \times \text{molality}
\][/tex]
Rearrange to solve for molality:
[tex]\[
\text{molality} = \frac{\Delta T_f}{K_f} = \frac{1.8^\circ \text{C}}{6.12 \, ^\circ \text{C} \cdot \text{kg/mol}} \approx 0.294 \, \text{mol/kg}
\][/tex]
4. Calculate Moles of Urea:
The molality ([tex]\(\text{molality}\)[/tex]) is the moles of solute per kilogram of solvent. Thus:
[tex]\[
\text{moles of urea} = \text{molality} \times \text{mass of solvent (kg)} = 0.294 \, \text{mol/kg} \times 0.1008 \, \text{kg} \approx 0.0296 \, \text{mol}
\][/tex]
5. Find the Mass of Urea:
The molar mass of urea, [tex]\((\text{NH}_2)_2\text{CO}\)[/tex], is approximately [tex]\(60.06 \, \text{g/mol}\)[/tex]. Therefore, the mass of urea is:
[tex]\[
\text{mass of urea} = \text{moles of urea} \times \text{molar mass of urea} = 0.0296 \, \text{mol} \times 60.06 \, \text{g/mol} \approx 1.78 \, \text{g}
\][/tex]
6. Present the Final Answer:
The mass of urea dissolved in the substance [tex]\(X\)[/tex] is approximately [tex]\(1.78 \, \text{g}\)[/tex].
1. Identify the Known Values:
- Normal freezing point of substance [tex]\(X\)[/tex]: [tex]\(-7.5^\circ \text{C}\)[/tex]
- New freezing point of the solution: [tex]\(-9.3^\circ \text{C}\)[/tex]
- Freezing point depression constant, [tex]\(K_f\)[/tex]: [tex]\(6.12 \, ^\circ \text{C} \cdot \text{kg/mol}\)[/tex]
- Mass of solvent, [tex]\(X\)[/tex]: [tex]\(100.8 \, \text{g} = 0.1008 \, \text{kg}\)[/tex]
2. Calculate the Freezing Point Depression:
[tex]\[
\Delta T_f = \text{Normal freezing point} - \text{New freezing point} = -7.5^\circ \text{C} - (-9.3^\circ \text{C}) = 1.8^\circ \text{C}
\][/tex]
3. Use the Formula for Freezing Point Depression:
[tex]\[
\Delta T_f = K_f \times \text{molality}
\][/tex]
Rearrange to solve for molality:
[tex]\[
\text{molality} = \frac{\Delta T_f}{K_f} = \frac{1.8^\circ \text{C}}{6.12 \, ^\circ \text{C} \cdot \text{kg/mol}} \approx 0.294 \, \text{mol/kg}
\][/tex]
4. Calculate Moles of Urea:
The molality ([tex]\(\text{molality}\)[/tex]) is the moles of solute per kilogram of solvent. Thus:
[tex]\[
\text{moles of urea} = \text{molality} \times \text{mass of solvent (kg)} = 0.294 \, \text{mol/kg} \times 0.1008 \, \text{kg} \approx 0.0296 \, \text{mol}
\][/tex]
5. Find the Mass of Urea:
The molar mass of urea, [tex]\((\text{NH}_2)_2\text{CO}\)[/tex], is approximately [tex]\(60.06 \, \text{g/mol}\)[/tex]. Therefore, the mass of urea is:
[tex]\[
\text{mass of urea} = \text{moles of urea} \times \text{molar mass of urea} = 0.0296 \, \text{mol} \times 60.06 \, \text{g/mol} \approx 1.78 \, \text{g}
\][/tex]
6. Present the Final Answer:
The mass of urea dissolved in the substance [tex]\(X\)[/tex] is approximately [tex]\(1.78 \, \text{g}\)[/tex].
