Question

how to prepare 1 Litre of 0.1 N HCL Solution

Answer 1

Answer:

Calculations: Stock bottle of 37% HCL. 37 ml of solute/100 ml of solution. Therefore add 8.3 ml of 37% HCL to 1 liter of D5W or NS to create a 0.1N HCL solution.

Answer 2

⇒Normality = 0.1 N

⇒Volume = 1 L

⇒N-FACTOR for HCl = basicity of HCl = no. of H+ ions produced from 1 molecule of HCl = 1

Normality = Molarity * N-Factor

0.1 = Molarity * 1

=> Molarity of HCl solution = 0.1 M (molar)

=> Molarity = (no. of moles of HCl)/(no.of litres of HCl solution)

=> 0.1 = (no. of moles)/1

=> no. of moles of HCl = 0.1 moles in 1 L of solution

Molar mass of HCl = 1 + 35.5 => 36.5 grams per mole.

=> So, 0.1 * 36.5

=> 3.65 grams of dry HCl gas should by dissolved in water to give 1 L of solution.

=> At, STP 1 mole of has 22.4 L of volume.

=> So, 0.1 moles of HCl at STP has volume = 0.1 * 22.4 => 2.24 L

=> So, 2.24 L of dry HCl at STP or 3.65 grams of dry HCl should be dissolved in water to give 1 L of 0.1 N HCl solution.