2.6g of an element X is reacted with an aqueous solution containing $NaOH$ and $N a N O_{3}$ to yield $N a_{2} X O_{2}$ and $N H_{3} . N H_{3}$ thus liberated is absorbed in $100 mL$ of $0.11 M H_{2} S O_{4}$ . The excess acid required $48 mL$ of $0.25 M NaOH$ for complete neutralization. Find the atomic weight of X.

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 65.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Milli equivalent of excess acid $= 48 \times 0.25 \times 1 = 12$

Milli equivalent of acid used $= 100 \times 0.11 \times 2 - 12$

$\begin{array}{l} = 10 \end{array}$

$= m mol$ of ${NH}_{3}$

$X + {OH}^{-} + {NO}_{3}^{-} ⟶ {XO}_{2}^{2 -} + {NH}_{3}$

$n. f = 2$ $n \cdot f = 8$

$∴$ Milli equivalent of ${NH}_{3} =$ Milli equivalent of X

$10 \times 8 = \frac{2.6}{\frac{A}{2}} \times 1000 \Rightarrow A = 65$

Watch 3-min video & get full concept clarity