Normality N of a solution is equal to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The normality of a solution is defined as the number of gram equivalents of a solute present per liter of solution. It is denoted by the symbol $N

Vedclass · Accepted Answer

$\frac{	ext{No. of gram equivalent of solute}}{	ext{Volume of solution in litre}}$

Vedclass · Answer

(B) The normality of a solution is defined as the number of gram equivalents of a solute present per liter of solution. It is denoted by the symbol $N$.$N = \frac{	ext{Number of gram equivalents of solute}}{	ext{Volume of solution in litres}}$Mathematically,it can be expressed as:$N = \frac{	ext{Weight of solute}}{	ext{Equivalent weight of solute}} 	imes \frac{1}{	ext{Volume of solution in litres}}$Where,$	ext{Equivalent weight} = \frac{	ext{Molar mass}}{n	ext{-factor}}$.

Normality $N$ of a solution is equal to

Similar Questions

Conc. $H_2SO_4$ has a density of $1.98 \ g/mL$ and is $98\%$ $H_2SO_4$ by weight. Its normality is ......... $N$.

The unit of mole fraction is:

$3.0 \ g$ ethanoic acid in $50 \ g$ benzene has . . . . . . molality. (Atomic weights: $H=1, C=12, O=16$).

Find out the molality of a solution prepared by dissolving $2.89 \ g$ of $NaCl$ in $0.159 \ L$ of water. The density of water is $1.00 \ g / mL$. (in $m$)

Molarity is expressed as

Vedclass Test Series

Exam Paper Generator

Online Exam Module