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What weight (in gg) of Na2CO3Na2CO3 of 80% purity would be required to neutralise 50mL50mL of 0.8MHCl0.8MHCl acid?

Asked by atharv2703123 6th February 2022, 4:39 PM
Answered by Expert
Answer:
Meq. of Na2CO3 = Meq. of H2SO4      ...... (For compltete neutralisation)
 
Meq. of Na2CO3 = 45.6 × 0.235
 
fraction numerator straight w over denominator 106 divided by 2 end fraction space cross times space 1000 space equals space 45.6 space space 0.235

straight w space equals space 0.5679 space straight g
 
Since, 95 g pure Na2CO3 is to be taken then weighed sample = 100 g
 
therefore space 0.5679 space straight g space pure space Na subscript 2 CO subscript 3 space is space to space be space taken space weighed space sample space equals space fraction numerator 100 cross times 0.5679 over denominator 95 end fraction space equals space 0.5978 space straight g
Answered by Expert 8th February 2022, 10:57 AM
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