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`        The mass of 70 percent pure H2SO4 required for neutralisation of 1 mol of NaOH -A 49 gB 98gC 70gD 34.3g `
one year ago

Arun
23342 Points
```							Dear Ayush If we take into context the neutralization reactionThen: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
```
one year ago
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