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The mass of 70 percent pure H2SO4 required for neutralisation of 1 mol of NaOH - A 49 g B 98g C 70g D 34.3g

The mass of 70 percent pure H2SO4 required for neutralisation of 1 mol of NaOH -
A 49 g
B 98g
C 70g
D 34.3g
 

Grade:12

1 Answers

Arun
25750 Points
5 years ago
Dear Ayush
 
If we take into context the neutralization reaction
Then:

h2so4 + 2NaOH ---> Na2SO4 + 2H2O
 
0.5 Moles of h2so4 are needed
 
And we know that the molar mass of h2so4 = 98g
 
0.5 moles mass of h2so4 would be = 98*0.5=49g
 
70 g h2so4 is there in 100 g solution (70% h2so4)
 
Thus 49g is there in 70g of solution
 
(49x100)/70=70g
 
Ans = 70 g
 

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