4.08 g of a mixture of BaO and unknown carbonate MCO 3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralisation. Identify the metal M.

Question

4.08 g of a mixture of BaO and unknown carbonate MCO 3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralisation. Identify the metal M. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Weight of mixture of BaO and M C O3=4.08 g On heating of M C O3 gives MO and C O2 Weight of B a O+M O=3.64 g Weight of CO 2 evolved =4.08-3.64=0.44 g Mols of CO 2=(0.44/44)=10-2 mol =10-2 × 103 =10 m mol =20 m Eq of CO 2 =20 m Eq MCO 3 Total acid =100 × 1=100 m Eq Excess of acid = Total acid - m" Eq of " N a O H =100-16 × 2.5=60 m Eq of excess acid (E w(B a O)=(M w/2)) m Eq of B a O=60-20=40 m Eq =((138+16) × 40/100 × 2)=3.08 g weight of M O=3.64-3.08=0.56 g 20 m " Eq of " M O=0.56 Equivalent weight of M O=0.56 × 1000=28 Molecular weight of M O=28 × 2=56 Hence, atomic weight of M=56-16=40.

Q. 4.08g of a mixture of BaO and unknown carbonate MCO3​ was heated strongly. The residue weighed 3.64g. This was dissolved in 100mL of 1NHCl. The excess acid required 16mL of 2.5NNaOH solution for complete neutralisation. Identify the metal M.

Q. $4.08 g$ of a mixture of $B a O$ and unknown carbonate $MC O_{3}$ was heated strongly. The residue weighed $3.64 g$ . This was dissolved in $100 m L$ of $1 N H Cl$ . The excess acid required $16 m L$ of $2.5 N N a O H$ solution for complete neutralisation. Identify the metal $M$ .