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(a) A solution is prepared by dissolving `3.65 g` of `HCl` in 500 mL of the solution. Calculate the normality of the solution (b) Calculate the volume of this solution required to prepare 250 mL of 0.05 N solution. |
Answer» Step I. Calculation of normality of solution Mass of `HCl = 3.65 g` Molar mass `HCl = 1+35.5 = 36.5 u = 36.5 "g mol"^(-1)` Basicity of `HCl` from the formula =1 `:.` Equivalent mass `HCl=("Molar mass")/("Bascity")=((36.5"g mol"^(-1)))/(1)=(("3.65 g"))/(("36.5 g equiv"^(-1)))=0.1 "equiv"` Volume of solution in litres `= 500 mL = (500)/(1000)=0.5 L` Normality of solution `(N)=("No. of equivalent of HCl")/("Volume of solution in litres")=("0.1 equiv.")/(("0.5 L"))="0.2 equiv L"^(-1)=0.2N` Step II. Calculation of volume of concentrated `HCl` solution required Normality of conc. `HCl = (N_(1))=0.2 N` Let the volume of conc. HCl needed `= V_(1)` Normality of dilute `HCl (N_(2))=0.05 N` Volume of dilute `HCl (V_(2))=250 mL` Applying normality equation : `N_(1)V_(1) -= N_(2)V_(2)` `(0.2 N)xxV_(1) = (0.05 N) xx ("250 mL")` `V_(1)=((0.05N)xx(250 mL))/((0.2 N))=62.5mL`. |
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