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`10ml` of sulphuric acid solution `(sp.gr.=1.84)` contains `98%` weight of pure acid. Calculate the volume of `2.5 M NaOH` solution required to just neutralise the acid. |
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Answer» Wt of solute `=10xx1.84xx(98)/(100)gm` So moles of solute `=(18.4)/(98)xx(98)/(100)=0.184` `n_(H+)=2xx0.184` `2xx0.184=(2.5xxV)/(1000)` `V=147.2` |
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