Aqueous solution of chloride of an element A containing 1 × 0-3 mol after complete hydrolysis required 30 mL of 0.1 M AgNO 3 solution to form AgCl. Element A is (Molar mass =133.5 g mol -1 )

Question

Aqueous solution of chloride of an element A containing 1 × 0-3 mol after complete hydrolysis required 30 mL of 0.1 M AgNO 3 solution to form AgCl. Element A is (Molar mass =133.5 g mol -1 ) (A) Al (B) P (C) Si (D) S

Tardigrade Admin · Accepted Answer

ACl n+n H 2 O arrow A ( OH )n+n HCl n HCl +n AgNO 3 arrow n AgCl +n HNO 3 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/284bbdd0115bfef5dc5d19263eab0509-.png" /> But AgNO 3 required =30 mL of 0.1 M =(30 × 0.1/1000)=3 × 10-3 mol Thus, 1 × 10-3 n=3 × 10-3 ∴ n=3 Thus, chloride is ACl 3 A+3 × 35.5=133.5 A=27 Thus, A is Al

Q. Aqueous solution of chloride of an element $A$ containing $1 \times 0^{- 3} m o l$ after complete hydrolysis required $30 m L$ of $0.1 M A g N O_{3}$ solution to form $A g Cl$ . Element $A$ is (Molar mass $= 133.5 g m o l^{- 1}$ )

$A l$

$P$

$S i$

$S$

Q. Aqueous solution of chloride of an element A containing 1×0−3mol after complete hydrolysis required 30mL of 0.1MAgNO3​ solution to form AgCl. Element A is (Molar mass =133.5gmol−1 )

Al

P

Si

S

ACln​+nH2​O→A(OH)n​+nHCl nHCl+nAgNO3​→nAgCl+nHNO3​ But AgNO3​ required =30mL of 0.1M =100030×0.1​=3×10−3mol Thus, 1×10−3n=3×10−3 ∴n=3 Thus, chloride is ACl3​ A+3×35.5=133.5 A=27 Thus, A is Al

Q. Aqueous solution of chloride of an element $A$ containing $1 \times 0^{- 3} m o l$ after complete hydrolysis required $30 m L$ of $0.1 M A g N O_{3}$ solution to form $A g Cl$ . Element $A$ is (Molar mass $= 133.5 g m o l^{- 1}$ )

$A l$

$P$

$S i$

$S$