A solution of urea received from some research laboratory has been marked mole fraction (x) and molality (m) at `10^@C`. While calculating its molality and mole fraction in the laboratory at `24^@C`, you will find

Options

A. Mole fraction (X) and molality (m)
B. Mole fraction (2X) and molality (2m)
C. Mole fraction (X/2) and molality (m/2)
D. Mole fraction (X) and molality (2m)

Answer

To solve the problem, we need to understand the concepts of mole fraction and molality, and how they are affected by temperature changes.

### Step-by-Step Solution:

1. **Understanding Mole Fraction and Molality**:
- **Mole Fraction (x)**: It is the ratio of the number of moles of a component to the total number of moles of all components in the solution. It is given by the formula:
\[ x = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
- **Molality (m)**: It is defined as the number of moles of solute per kilogram of solvent. The formula is:
\[ m = \frac{n_{solute}}{mass_{solvent \, (kg)}} \]

2. **Temperature Independence**:
- Both mole fraction and molality are considered temperature-independent properties. This means that changes in temperature do not affect their values.

3. **Given Conditions**:
- The solution of urea is marked with mole fraction (x) and molality (m) at 10°C.
- We need to find the values of mole fraction and molality at 24°C.

4. **Conclusion**:
- Since mole fraction and molality are temperature-independent, the values calculated at 10°C will remain unchanged when calculated at 24°C.
- Therefore, the mole fraction (x) and molality (m) at 24°C will be the same as those at 10°C.

### Final Answer:
- The mole fraction (x) and molality (m) at 24°C will be the same as at 10°C.

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