Question

12. Regis Compound Normality Gram molecular weight Volume Grams (or milligrams) 10 ml 200 ml MgCl2 Ca(NO3)2 HCI NaOH Н;РО, 400mg 500 g Se 40L 4000g 978

Answer 1

Compound Normality regel 14 Grams for miligrains) I 1.904 g 16.4088 g 1.823 g 1599.88 g "Gram Molecult Volume ar weight 195.211gmol some 164.088 gland 200me 36 46 g/mol 10mL 39.997 glamol 40L 97.994 g/mol some Ca(NO3)2 HCL NAOH 979.94 mg H3PO4 for Normality = Valency x Molarity | & Molecular inright of gel2 = 95.218 gimole Valency of reght is = 2 This 4 Normal sortion = 2 molar Solution. 1 Molar solution contains a 95.21ag of ugch in 1000 . 10me solution of 2M solution will have - 2x 951211 X10wLxg 1000 mL C = 1.904 g hat Ca(NO3), solution have Normality = 1. i Molarity, a 0.5M as Ca2+ ion has valency=2 Oohh of Ca(NO3), solution contains (16470 88 X0.5 g im 1000me = 82.04 again 1000 me in 200ml solition contains - 820044 g x 200m 1000 ml = 16.4088 g

or Ha Solution, Normality Molarity. Thus 5N sointion = 543 s 5N solution = 5M solution of Ha 5M svention of Hel= (5x36468) of Hel in 1000 na = 182.3 g of Hell in 1000 ml on some solution will - 182.3 g x 10 ml contain =- 1000 ml = 1.823 g (1) for NaoH Solutions Normality Molarity This In Solntion = IM Sointion of Nadh IM solution of NaoH Contams 59. OH in 1000 ml . 40L solution will contain = 39.9978x40 L 1000 me 2 39,997gx404 = 1599.88 g For H₂POg Solnition, Normality = 3x Molarity, as valency of Porn is 3 This 3 Normal H3PO4 solution Imolar solution = IM solution of H₂PO4 IM solution of Hy POA will have 97.994g im 1ooo me some solution contains = 97.994 g x lowe 1000 mh = 0.97994 g = 979.99 mg