How many molecules of water are present as water of crystallisation in Borax `Na_(2)B_(4)O_(7).nH_(2)O` if it loses `47.117%` of mass on heating till it becomes anhydrous?

Options

A. 10
B. 8
C. 6
D. 4

Answer

To determine how many molecules of water are present as water of crystallization in Borax \( \text{Na}_2\text{B}_4\text{O}_7 \cdot n\text{H}_2\text{O} \), given that it loses \( 47.117\% \) of its mass on heating, we can follow these steps:

### Step 1: Assume the Mass of Borax
Assume the total mass of the hydrated borax \( \text{Na}_2\text{B}_4\text{O}_7 \cdot n\text{H}_2\text{O} \) to be \( 100 \, \text{g} \).

### Step 2: Calculate the Mass of Water Lost
Since it loses \( 47.117\% \) of its mass upon heating, the mass of water lost can be calculated as:
\[ \text{Mass of water} = 100 \, \text{g} \times \frac{47.117}{100} = 47.117 \, \text{g} \]

### Step 3: Calculate Moles of Water
To find the number of moles of water, use the molar mass of water \( \text{H}_2\text{O} \), which is \( 18 \, \text{g/mol} \):
\[ \text{Moles of water} = \frac{47.117 \, \text{g}}{18 \, \text{g/mol}} \approx 2.61 \, \text{moles} \]

### Step 4: Calculate the Mass of Anhydrous Borax
The remaining mass after losing water is:
\[ \text{Mass of } \text{Na}_2\text{B}_4\text{O}_7 = 100 \, \text{g} - 47.117 \, \text{g} = 52.883 \, \text{g} \]

### Step 5: Calculate Moles of Anhydrous Borax
Next, we need to find the molar mass of anhydrous borax \( \text{Na}_2\text{B}_4\text{O}_7 \):
- Sodium (Na): \( 23 \, \text{g/mol} \times 2 = 46 \, \text{g} \)
- Boron (B): \( 11 \, \text{g/mol} \times 4 = 44 \, \text{g} \)
- Oxygen (O): \( 16 \, \text{g/mol} \times 7 = 112 \, \text{g} \)

Total molar mass of \( \text{Na}_2\text{B}_4\text{O}_7 \):
\[ \text{Molar mass} = 46 + 44 + 112 = 202 \, \text{g/mol} \]
Now calculate the moles of \( \text{Na}_2\text{B}_4\text{O}_7 \):
\[ \text{Moles of } \text{Na}_2\text{B}_4\text{O}_7 = \frac{52.883 \, \text{g}}{202 \, \text{g/mol}} \approx 0.261 \, \text{moles} \]

### Step 6: Relate Moles of Water to Moles of Anhydrous Borax
From the formula \( \text{Na}_2\text{B}_4\text{O}_7 \cdot n\text{H}_2\text{O} \), we know that 1 mole of \( \text{Na}_2\text{B}_4\text{O}_7 \) corresponds to \( n \) moles of water. Therefore:
\[ 0.261 \, \text{moles of } \text{Na}_2\text{B}_4\text{O}_7 = n \times 0.261 \, \text{moles of water} \]
Setting this equal to the moles of water calculated:
\[ 0.261n = 2.61 \]
Solving for \( n \):
\[ n = \frac{2.61}{0.261} \approx 10 \]

### Conclusion
Thus, the number of molecules of water present as water of crystallization in Borax is \( n = 10 \).

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