Question

\( x \) g of an organic compound was analysed by Kjeldahl’s method for the estimation of nitrogen. The ammonia gas evolved from the compound was passed into 100 mL of 0.2 M H\(_2\)SO\(_4\). The excess of acid required 40 mL of 0.5 M NaOH solution for complete neutralisation. What is \( x \) when the percentage of nitrogen in it is 46%?

• A. 0.508 g
• B. 0.408 g
• C. 0.608 g
• D. 0.808 g

Hint:

The Kjeldahl method is effective for determining the nitrogen content in organic compounds by converting all nitrogen into ammonia.

Answer 1

The Correct Option is C

Step 1: Determining the amount of acid neutralized by ammonia gas.

Moles of H\(_2\)SO\(_4\) taken:

\[ \text{Moles} = M \times V = 0.2 \times \frac{100}{1000} = 0.02 \, \text{moles} \]

Moles of NaOH used to neutralize excess H\(_2\)SO\(_4\):

\[ \text{Moles} = 0.5 \times \frac{40}{1000} = 0.02 \, \text{moles} \]

Since 2 moles of NaOH neutralize 1 mole of H\(_2\)SO\(_4\), moles of unreacted H\(_2\)SO\(_4\) are:

\[ \frac{0.02}{2} = 0.01 \, \text{moles} \]

Moles of H\(_2\)SO\(_4\) reacted with NH\(_3\):

\[ 0.02 - 0.01 = 0.01 \, \text{moles} \]

Moles of nitrogen in NH\(_3\):

\[ 2 \times 0.01 = 0.02 \, \text{moles} \] Step 2: Calculating the mass of nitrogen.

Mass of nitrogen:

\[ 0.02 \times 14 = 0.28 \, \text{g} \]

Given that nitrogen is 46% of the organic compound:

\[ x = \frac{0.28}{0.46} = 0.608 \, \text{g} \]

Thus, the value of \( x \) is 0.608 g.