A 6.90 M solution of kOH in water contain 30% by weight of KOH.What is the density of the solution?

A 30 % by weight KOH solution must contain 30 g of solid KOH and 70 g water (density of water is 1 g/cc) ;

Molecular mass of KOH = 56 ;

In a 6.90 M KOH solution (56 X 6.90) g KOH dissolved in 1000 mL solution ;

In a 6.90 M KOH solution 30 g KOH is dissloved in (30 X 1000)/(56 X 6.90) mL or in

77.64 mL solution ;

Amount of water present in that solution dissolving 30 g KOH = 70 g (since it is a 30 % by weight solution) ;

density = mass / volume

= (30 + 70) / 77.64 g/mL = 1.288 g/mL (or g/cc) .

This is just what I would do, I'm not fully confident in my answer here.

You can choose any amount volume you like, for ease I'm going to use 1L

n=CV

n(KOH)= 6.90x1

n(KOH)=6.90

nM=m

6.90x(39.1+16+1)=387g

Multiply that by 30% because of 30% by weight

116.127g of KOH in 1L of solution

D=m/V

D(KOH)=116.127/1

D(KOH)=116.127g/L

I'm actually really not to sure about this, but if I got ask this question, this is how I would approach it. If you have answers it would be great to list it.