Question

what mass of 80% NaOH required to neutralize 250 gm 98% of pure sulphuric acid​

Answer 1

Given : 10% pure 100ml, 1MH

2

SO

4

sol

: 80% pure NaOH

To find : at of NaOH required for neutralization A/C

As 10% pure 100ml, 1M.H

2

SO

4

∴ mole of H

2

SO

4

=

100

10

×

1000

100

×1=0.01 mole H

2

SO

4

Now 2 mole NaOH required for neutralization of 1 mole H

2

SO

4

.

∴ mole of NaOH required =2×0.01

=0.02moleNaOH

∴ at of NaOH =0.02×40=0.8gmNaOH

as it is 80% pure

∴ mass required ×

100

80

=0.8gm

∴NaOH required ×

80

0.8

=1gm

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