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

Explanation

6.90 M solution of KOH contains 6.90 moles of KOH in 1000 mL of solution.
Wt. of KOH in solution `= 6.90 xx 56 = 386.4 g`
Wt of KOH in 1000 mL solution = 386.4 g
Since the solution is 30% by weight , it means that 30 g of KOH are present in 100g of solution .
`therefore ` 386.4 of KOH is present in ` = (100)/(30) xx 386.4`
= 1288 g of solution .
Density ` = ("Weight")/("Volume")`
` = (1288)/(1000) = 1.288 g mL^(-1)`