$\begin{array}{1 1}1.288gcm^{-3}\\2.288gcm^{-3}\\3.288gcm^{-3}\\4.288gcm^{-3}\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

A 6.90M solution contains 6.90 mol KOH in 1000 $cm^3$ of the solution

Mass of 6.90 mol of KOH =$6.90\times 56$

$\Rightarrow 386.4g$

A 30% solution contains 30g of KOH present in 100g of solution.

$\therefore 386.4$ g of KOH is present in = $\large\frac{100\times 386.4}{30}$

$\Rightarrow 1288g$ of solution

Density of KOH solution =$\large\frac{\text{Mass}}{\text{Volume}}$

$\Rightarrow \large\frac{1288g}{1000cm^3}$

$\Rightarrow 1.288g cm^{-3}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...