0.5 gm of an organic substance containing prosphorous was heated with sonc. `HNO_3` is the carius tube. The phosphoric acid thus formed was precipitated with magnesia mixture `(MgNH_4PO_4)` which on ignition gave a residue of 1.0 gm of magnesium phyrophosphate `(Mg_2P_2O_7)`. The precentage of phosphorous in the organic compound is:

Options

A. `55.85%`
B. `29.72%`
C. `19.18%`
D. `20.5%`

Explanation

To find the percentage of phosphorus in the organic compound, we can follow these steps:

### Step 1: Understand the Reaction
When the organic substance containing phosphorus is heated with concentrated nitric acid (HNO₃), it forms phosphoric acid (H₃PO₄). The phosphoric acid is then precipitated with a magnesium mixture to form magnesium pyrophosphate (Mg₂P₂O₇).

### Step 2: Determine the Molar Masses
We need to know the molar masses of the compounds involved:
- Molar mass of magnesium pyrophosphate (Mg₂P₂O₇):
- Mg: 24.31 g/mol (2 Mg = 2 × 24.31 = 48.62 g/mol)
- P: 30.97 g/mol (2 P = 2 × 30.97 = 61.94 g/mol)
- O: 16.00 g/mol (7 O = 7 × 16.00 = 112.00 g/mol)

Total molar mass of Mg₂P₂O₇ = 48.62 + 61.94 + 112.00 = 222.56 g/mol (approximately 222 g/mol for calculation)

- Molar mass of phosphorus (P): 30.97 g/mol

### Step 3: Calculate the Weight of Phosphorus in Magnesium Pyrophosphate
From the formula of magnesium pyrophosphate (Mg₂P₂O₇), we can see that it contains 2 moles of phosphorus. Thus, the weight of phosphorus in 1 mole of Mg₂P₂O₇ is:
- Weight of phosphorus in Mg₂P₂O₇ = 2 × 30.97 g/mol = 61.94 g/mol (approximately 62 g/mol for calculation)

### Step 4: Set Up the Percentage Calculation
The percentage of phosphorus in the organic compound can be calculated using the formula:
\[ \text{Percentage of P} = \left( \frac{\text{Weight of P in Mg}_2\text{P}_2\text{O}_7}{\text{Weight of Mg}_2\text{P}_2\text{O}_7} \right) \times \left( \frac{\text{Weight of Mg}_2\text{P}_2\text{O}_7 \text{ formed}}{\text{Weight of organic compound}} \right) \times 100 \]

### Step 5: Substitute the Values
Given:
- Weight of Mg₂P₂O₇ formed = 1.0 g
- Weight of organic compound = 0.5 g

Substituting the values into the formula:
\[ \text{Percentage of P} = \left( \frac{62}{222} \right) \times \left( \frac{1.0}{0.5} \right) \times 100 \]

### Step 6: Simplify the Calculation
Calculating the above expression:
1. Calculate \( \frac{62}{222} \approx 0.279 \)
2. Calculate \( \frac{1.0}{0.5} = 2 \)
3. Now, multiply: \( 0.279 \times 2 \times 100 = 55.85 \)

### Final Answer
The percentage of phosphorus in the organic compound is approximately **55.85%**.

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