Question

0.1 g of metal carbonate is dissolved in 20 ml 0.1 N sulphuric acid what Is the eq.wt of metal carbonate​

Answer 1

Answer:

त्ह्रल्लगर्ेाग ररिहरमरगगीगकमूिगमगगिगीhallo

Answer 2

Answer:

we have,

MCO

3

+2HCl→MCl

2

+CO

2

+H

2

O

we have, 50 ml of 1 N HCl

no . of moles of HCl =

1000

1∗50

= 0.05 mole

100 ml of 0.1 N NaOH , have no of moles = 0.01mole

moles of HCl left after neutralisation with NaOH = 0.05 - 0.01 = 0.04 mole

the remaining moles of HCl is reacting with metal carbonate

2 mole of HCl is reacting with 1 mole of metal carbonate

so, 0.04 mole of HCl is reacting with metal carbonate =

2

0.04

= 0.02 moles

moles of metal carbonate = 0.02 =

molecularweight

givenweight

we get, molecular weight = 100g

B) we have, N

1

V

1

=N

2

V

2

where, V

2

is final volume

so we get, 3.

1000

75

=V

2

∗0.1

V

2

= 225 ml

amount of water added = 225 - 75 = 145 ml

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