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10ml of sulphuric acid solution (sp.gr.=...

`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.

Text Solution

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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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