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A 6.90 M solution of KOH contains 30% by...

A `6.90 M` solution of `KOH` contains 30% by weight of `KOH`. Calculate the density of the solution.

Text Solution

Verified by Experts

Let `V=1lt`, then moles of solute`=6.9`
wt of solute `=6.9xx56gm`
`%=(M_("solute"))/(M_("solute"))xx100` So, `=(6.9xx56)/(d xx1000)xx100=30., d=1.288 gm//"litre"`
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