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Question

When 4.08 g of a mixture of BaO and unknown carbonate MCO3 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
Chromium
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B
Nickel
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C
Zinc
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D
Calcium
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Solution

The correct option is D Calcium
Weight of MCO3 and BaO=4.08g (given)
Weight of residue =3.64g (given)
Weight of CO2 evolved on hating =(4.083.64)g=0.44g
=0.44/44=0.01mole
Number of moles of MCO3=0.01mole
[MCO3heatMO+CO2]
volume of 1NHCl in which residue is dissolved =100ml
volume of 1NHCl used for dissolution =(1002.5×16)ml=60ml
=601000=0.06 equivalents
The chemical equation for dissolution can be written as
BaO+MOResidue+4HClBACl2+MCl2+2H2O
[Number of moles of BaO and MO=1+1=2]
Number of moles of BaO+ Number of moles of MO=0.062=0.03
Number of moles of BaO=(0.030.01)=0.02 moles
Molecular weight of BaO=138+16=154
Weight of BaO=(0.02×154)g
=30.8g
Weight of MCO3=(4.083.08)=1.0g
Since weight of 0.01 mole of MCO3=1.0g
Mol. wt. of MCO3=10.01=100
Hence atomic weight of unknown M=(10060)=40
The atomic weight of metal is 40 so the metal M is Ca

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