`DeltaS` for freezing of 10 g of `H_(2)O(l)` (enthalpy of fusion is 80 cal/g) at `0^(@)C` and 1 atm is :

Options

A. `12.25 J//K`
B. `-0.244 J//K`
C. `-2.93 J//K`
D. `-12.25 J//K`

Explanation

To calculate the change in entropy (ΔS) for the freezing of 10 g of water (H₂O) at 0°C and 1 atm, we can follow these steps:

### Step 1: Identify the given data
- Mass of water (m) = 10 g
- Enthalpy of fusion (ΔH_fusion) = 80 cal/g
- Temperature (T) = 0°C = 273 K (convert Celsius to Kelvin by adding 273)

### Step 2: Determine the enthalpy change for freezing
Since freezing is the reverse of melting, we take the negative of the enthalpy of fusion:
- ΔH_freezing = -ΔH_fusion = -80 cal/g

### Step 3: Calculate the total enthalpy change for 10 g of water
Now, we can calculate the total enthalpy change for 10 g of water:
- Total ΔH_freezing = ΔH_freezing × mass = -80 cal/g × 10 g = -800 cal

### Step 4: Convert the enthalpy change from calories to joules
To convert calories to joules, we use the conversion factor:
1 cal = 4.184 J
- Total ΔH_freezing in joules = -800 cal × 4.184 J/cal = -3347.2 J

### Step 5: Calculate the change in entropy (ΔS)
The change in entropy is given by the formula:
\[ \Delta S = \frac{\Delta H}{T} \]
Substituting the values we have:
- ΔS = Total ΔH_freezing / T = -3347.2 J / 273 K

### Step 6: Perform the calculation
Now, we can calculate ΔS:
- ΔS = -3347.2 J / 273 K ≈ -12.25 J/K

### Final Answer
Thus, the change in entropy (ΔS) for the freezing of 10 g of water at 0°C is approximately -12.25 J/K.

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