Question

In phosphorus estimation, 0.5 g of an organic compound gives 0.75 g of \( \mathrm{Mg_2P_2O_7} \). The percentage of phosphorus (P) in the compound is (nearest integer):

Hint:

In gravimetric estimation problems, always remember: \[ %\text{Element} = \frac{\text{Mass of precipitate} \times \text{Mass of element in 1 mole of precipitate}} {\text{Molar mass of precipitate} \times \text{Mass of sample}} \times 100 \] Round off only at the final step if the question asks for the nearest integer.

Answer 1

Correct Answer: 42

To determine the percentage of phosphorus (P) in the compound, follow these steps:

Step 1: Calculate Moles of \( \mathrm{Mg_2P_2O_7} \)
The molar mass of \( \mathrm{Mg_2P_2O_7} \) is calculated as follows:
\( \mathrm{Mg} = 24.305 \, \text{g/mol} \), \( \mathrm{P} = 30.974 \, \text{g/mol} \), \( \mathrm{O} = 15.999 \, \text{g/mol} \).
Molar mass of \( \mathrm{Mg_2P_2O_7} = 2 \times 24.305 + 2 \times 30.974 + 7 \times 15.999 = 222.548 \, \text{g/mol} \).
Given \( 0.75 \, \text{g of} \, \mathrm{Mg_2P_2O_7} \), the moles are:
\(\frac{0.75}{222.548} = 0.00337 \, \text{mol}\). 

Step 2: Relate moles of \( \mathrm{Mg_2P_2O_7} \) to phosphorus
1 mole of \( \mathrm{Mg_2P_2O_7} \) contains 2 moles of phosphorus.
Thus, 0.00337 moles of \( \mathrm{Mg_2P_2O_7} \) contain \( 2 \times 0.00337 = 0.00674 \, \text{mol} \, \text{of P} \).

Step 3: Calculate the mass of phosphorus
\( \text{Mass of P} = 0.00674 \times 30.974 = 0.2087 \, \text{g} \).

Step 4: Calculate percentage of phosphorus in the compound
\(\text{Percentage of P} = \left(\frac{0.2087}{0.5}\right) \times 100 = 41.74\%\).
Rounding to the nearest integer, the percentage of P is \( 42\% \).

The computed percentage \(42\%\) matches the expected range of 42 to 42, confirming the accuracy.