Correct Answer: The correct answer is 41.61%.

Step-by-step Explanation:

In the estimation of phosphorus in an organic compound (Carius method), phosphorus present in the organic compound is converted into magnesium pyrophosphate (${\text{Mg}}_2{\text{P}}_2{\text{O}}_7$).

1. Molar Masses:

Molar mass of Phosphorus ($\text{P}$) = 31 g/mol

Molar mass of Magnesium ($\text{Mg}$) = 24 g/mol

Molar mass of Oxygen ($\text{O}$) = 16 g/mol

Molar mass of ${\text{Mg}}_2{\text{P}}_2{\text{O}}_7$:

$\text{Molar mass}=\; (2\; \times \; 24)\; +\; (2\; \times \; 31)\; +\; (7\; \times \; 16)\; =\; 48\; +\; 62\; +\; 112\; =\; 222g/mol$

2. Mass of Phosphorus in ${\text{Mg}}_2{\text{P}}_2{\text{O}}_7$:

1 mole of ${\text{Mg}}_2{\text{P}}_2{\text{O}}_7$ (222 g) contains 2 moles of Phosphorus, which equals $2\times 31=62g$ of phosphorus.

3. Formula for Percentage of Phosphorus (% P):

$\text{\% of P}=\frac{62}{222}\times \frac{\text{Mass of}{\text{Mg}}_2{\text{P}}_2{\text{O}}_7}{\text{Mass of organic compound}}\times 100$

4. Calculation:

Given data:

Mass of organic compound = 1 g

Mass of ${\text{Mg}}_2{\text{P}}_2{\text{O}}_7$ = 1.49 g

Substitute the given values into the formula:

$\text{\% of P}=\frac{62}{222}\times \frac{1.49}{1}\times 100$

$\text{\% of P}=\frac{92.38}{222}\times 100=0.416126\times 100\approx 41.61\%$

Hence, the percentage of phosphorus in the organic compound is 41.61%.