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What volume of 0.25 `MH_2SO_4` is required to neutralise 1.90 g of a mixture containing equimolar amounts of `NaHCO_3` and `NaCO_3`? |
Answer» x mol each of `NaHCO_3` and `Na_2CO_3` `implies84x+106x=1.9g` (given) `x=0.01` `implies` milli" mol of "each `=10` Now m" Eq of "`H_2SO_4=m" Eq of "(NaHCO_3+Na_2CO_3)` `2xx0.25xxV=1xx10+2xx10` `impliesV=60mL` |
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