Ask Questions, Get Answers

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

A sample of sulphuric acid is 98% $H_2SO_4$ by weight (density =$1.10gmL^{-1})$.Calculate normality

$\begin{array}{1 1}22N\\42N\\62N\\12N\end{array} $

Can you answer this question?

1 Answer

0 votes
98% $H_2SO_4$ solution means that 98 g of $H_2SO_4$ are present in 100g of solution in water
Moles of $H_2SO_4=\large\frac{98}{98}$$=1$
Volume of solution =$\large\frac{\text{Mass}}{\text{Density}}$
$\Rightarrow \large\frac{100}{1.10}$$=90.9mL$
Normality =$\large\frac{\text{Gram equivalents of solute}}{\text{Volume of solution}}$$\times 1000$
Equivalent mass of $H_2SO_4=\large\frac{\text{Molar mass}}{2}$
$\Rightarrow \large\frac{98}{2}$$=49$
Basicity of $H_2SO_4=2$
Gram equivalents of $H_2SO_4=\large\frac{\text{Mass of }H_2SO_4}{\text{equivalent mass}}$
$\Rightarrow \large\frac{98}{49}=$$2$
Normality =$\large\frac{\text{Gram equivalents of }H_2SO_4}{\text{Volume of solution(mL)}}$$\times 1000$
$\Rightarrow \large\frac{2\times 1000}{90.9}$$=22N$
answered Aug 6, 2014 by sreemathi.v

Related questions

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