Question

25ml,0.1N H2SO4 nutralised with 2ml XN sodium carbonate what will be the g/l sodium carbonate​

Answer 1

Answer:

hey mate here is your answer

The specific gravity of sodium carbonate is 1.25 g/mL.

Hence, the mass of 25 mL of solution of sodium carbonate is 1.25×25=31.25 g.

The mass of HCl used for neutralization is 100032.9×109.5=3.60 g.

The number of moles of HCl in 3.60 g is 36.53.60=0.0987.

2 moles of HCl will neutralize 1 mole of sodium carbonate.

The number of moles of sodium carbonate neutralized by 0.0987 moles of HCl are 20.987=0.04935

Thus, 31.25 g of sodium carbonate contains 0.04935 moles.

Hence, 125 g of sodium carbonate will contain 31.25125×0.04935=0.1974 moles.

They will neutrlaize  0.1974 moles of sulphuric acid which corresponds to 2×0.1974=0.3948 g eq of sulphuric acid.

The volume of 0.84 N sulphuric acid required will be 0.840.3948=0.470 L or 470 mL.

please mark me as a brainliest