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

### Step 1: Assume the mass of Borax
Assume the mass of \( \text{Na}_2\text{B}_4\text{O}_7 \cdot n\text{H}_2\text{O} \) is \( 100 \) grams.

### Step 2: Calculate the mass of water lost
The mass of water lost during heating is given by:
\[ \text{Mass of water} = 100 \times \frac{47.117}{100} = 47.117 \text{ grams} \]

### Step 3: Calculate the mass of anhydrous Borax
Now, subtract the mass of water from the total mass to find the mass of anhydrous Borax:
\[ \text{Mass of anhydrous Borax} = 100 - 47.117 = 52.883 \text{ grams} \]

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

Adding these together:
\[ \text{Molar mass of } \text{Na}_2\text{B}_4\text{O}_7 = 46 + 44 + 112 = 202 \, \text{g/mol} \]

### Step 5: Calculate the moles of anhydrous Borax
Now, calculate the moles of anhydrous Borax:
\[ \text{Moles of } \text{Na}_2\text{B}_4\text{O}_7 = \frac{52.883 \text{ g}}{202 \text{ g/mol}} \approx 0.261 \text{ moles} \]

### Step 6: Calculate the moles of water
We know that the moles of water lost (as \( n \) moles) can be calculated from the moles of water:
\[ \text{Moles of water} = \frac{47.117 \text{ g}}{18 \text{ g/mol}} \approx 2.61 \text{ moles} \]

### Step 7: Relate moles of Borax to moles of water
From the reaction, we can say:
\[ n \times 0.261 = 2.61 \]
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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