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Q.

The mass of 80% pure H2SO4 required to completely neutralise 60g of NaOH is

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a

92g

b

58.8g

c

73.5g

d

98g

answer is A.

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Detailed Solution

large mathop {mathop {{H_2}S{O_4}}limits_{98g} }limits_{'X';g}^{1;mole} ; + mathop {mathop {2NaOH}limits_{2 times 40g} }limits_{60g}^{2;mole} ; to ;N{a_2}S{O_4}; + ;2{H_2}O

 


 

large X; = ;frac{{60 times 98}}{{2 times 40}}; = ;frac{{49 times 3}}{2}g

 


 

large ;frac{{49 times 3}}{2}g

 

of 100%H2SO4 is required but the purity is only 80%
Wt of 80% pure H2SO4 required =

large frac{100}{80}times frac{49times 3}{2}simeq 92g
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