Question

If 6 M $KOH$ solution is 28% by weight, then what is the specific gravity of this solution?

• A. $1.2$
• B. $2.4$
• C. $0.6$
• D. $1.8$

Answer 1

The correct option is A $1.2$

28% $\left(\omega /\omega \right)$ means 28 gm KOH in 100 gm solution.

$Molarity=\frac{\omega }{M\omega t}\times \frac{1000}{V\left(ml\right)}$

$6=\frac{28}{56}\times \frac{1000}{V\left(ml\right)}$

$V=\frac{1000}{12}=83.33ml$

$Density=\frac{100}{83.33}$

$=1.2g/ml$

$Specificgravity=1.2g/ml$