Q.

One mole of Sodium bicarbonate on strong heating suffers a loss in weight of

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a

31g

b

62g

c

44g

d

22g

answer is A.

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Detailed Solution

Weight lost =  Weight of gas liberated out;Temperature required to decompose NaHCO3 is so high that water also is liberated out as vapour ;2NaHCO3s →∆ Na2CO3s+CO2g+H2Og;   2moles                                         44g          18g........62g(total)   1mole                                           22g            9g........31g(total)
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