Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
0 votes

A 6.90 M solution of KOH in water contains 30% by mass of KOH.Calculate the density of the KOH solution ?[Molar mass of KOH=$56g mol^{-1}$]

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

Can you answer this question?

1 Answer

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 =$\large\frac{\text{Mass}}{\text{Volume}}$
$\Rightarrow \large\frac{1288g}{1000cm^3}$
$\Rightarrow 1.288g cm^{-3}$
answered Aug 8, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App