Upon heating a litre if N/2HCl solution,2.675g HCl is lost and the volume of the solution shrinks to 750ml.calculate....i)Normality of resulting solution. ii)number of milliequivalents of HCl in 100ml of final solution..........pls give detaild explanation ....
Answers
Answer:
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Explanation:
An solution of HCl means it has normality equal to its molarity since it has 1 equivalent proton.
Original solution of Hydrochloric Acid is 12.178 N (it is also 12.178 M) for a density of 1.2 g/mL, formula weight of 36.46 g/mol, and concentration of 37% w/w for 1 L of HCl acid.
Since it is 37% w/w, solution, it contains 370 g HCl in 1000 g (nearly 1 L for water)
Now, N/2 (0.5 N) HCl is made by 41.059 mL from original solution. i.e. 1000 mL of 0.5 N HCl has = 37100×41.059=15.192 g HCl37100×41.059=15.192 g HCl
Now, the question is heating this solution so that 2.675 g of HCl is evaporated. So, remaining vol = 15.192 - 2.675 = 12.517 g.
And, the volume is 570 mL. So, Molarity = moles of solute/ L of solution = 12.517570×36.5×100012.517570×36.5×1000 = 0.6 M.
Since normality = molarity for HCl, So, it is 0.6N.
For monovalent ions, 1 meq = 1 mmol, So, meq of 0.5 N (0.5 M) HCl solution = 0.5 x 1000 = 500 (since 1 mol =1000 mmol)