Question

I need help with questions 12 and 13 please.

12 (5 points) How much 2.5N solution of HsAso, can I make with 57 mL of stock solution that is 84% assay and 1.14 specific gravity? 13 (5 points) If we have a 4N solution of HCl that has 0.03645g of HCl in the solution, how many microliters of solution do we have?

Answer 1

Volume of solution = 57 ml, specific gravity= 1.14 density = specific gravity of substance x density of water= 1.14*1=1.14g/cc mass = density* specific gravity = 57*1.14=65 gm it's assay is 84% mass of h3aso4= 65*.84 =54.6 gmsince there are 3 replaceable hydrogen ions in heaso4 it's equivalent weight = molar mass/3 =142/3=47.3 g/mole , normality = gmequivalents/ liter gm equivalents = mass/ equivalent weight =54.6/47.3=1.155 , 2.5 gm equivalents are there in 1L, i.e 1000ml 1.155 gm equivalents are there in 1.155*1000/2.5=462 ml

2. Eq.weight of hcl= 36.5

Gm equivalents in .03645gm of hcl= 0.03645/36.5= .0099

2.5 gm equivalents are there in 1L

.0099 gm equivAlents in 0.0099/2.5=.000399L=1000000*0.000399 microliters =399 microliters

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