Question

Write down the steps to prepare 0.1NHCl.

Answer 1

Solution:

Step:1: Molarity of Hydrogen chloride $\left(\mathrm{HCl}\right)$

1. Volume of $\mathrm{HCl}$$=37\%$v/v or $36.5\mathrm{g}\mathrm{of}$ $\mathrm{HCl}$ per$100\mathrm{g}$of solution
2. Specific gravity$=1.19\mathrm{g}/\mathrm{ml}$
3. $370\mathrm{ml}/1000\mathrm{ml}\times 1.19\mathrm{g}/\mathrm{ml}$$=440.3\mathrm{g}/\mathrm{L}$
4. $\mathrm{HCl}$ Molecular weight$=36.5$
5. Molarity=$\frac{\mathrm{mass}\mathrm{of}\mathrm{solute}}{\mathrm{molar}\mathrm{mass}\mathrm{of}\mathrm{solution}}$
6. Molarity=$\frac{440.3\mathrm{g}}{36.5\mathrm{g}}=12.06\mathrm{M}$$\equiv 12\mathrm{M}$

Step:2:Normality of $\mathrm{HCl}$

1. To find normality: ${\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2$
2. In the equation, N=Normality of solution and V=Volume of solution
3. Applying the above Formula,$12\times {\mathrm{V}}_1=0.1\times 1$
4. V₁₌$8.33\mathrm{ml}$

If we add $8.33\mathrm{ml}$ in$1000\mathrm{l}$ the water we get $0.1\mathrm{N}$$\mathrm{HCl}$