1.

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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