For Unit For Quantities And Concentrations, Exercises of Analytical Chemistry

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Methanol has a density of 2. 79 g/mL. If 16. 5 mL of this

methanol is added to water to make a 2000. 0 mL solution,

calculate the concentration of the prepared methanol solution

in,

i) ppm

ii) molarity

CHOH = (12)^ +^3

• 16+1=

• · 2 L Parts (^) per million^ (ii)molarity mularity = Pmoc (^) sacute (mma/mor) volume solution^ (ML/2) 2 i) (^) I 2.79 (^) g/mL = 16.5 mL 1.4386 mol = (^) 46. g NcrsOn =46035g M^ = 22 32 mm PPM =

= 23817.5ppon => 1.41386 mol

0.7193M/

2000m2x

A 500 mL solution contains 26.3 g FeCl 3

. Calculate the i) concentration of Fe 3+ in g/L. ii) molarity of Cl

. -^ 500mL I = (^) 0.5L Concentration 9 -0558g Fest = 0- = 18.1g / 2 I (^) ) (^) need , = 26- (55-85)+(35.45×3) (^) ii ) (^) molarity = Mot Homme = (^0) -1621 MOI MOI CI^ =^3 ✗ MOI Feel} M^ =^ 0.483 MOI MOI Fest^ = (^1) ×0-1625 MOI (^) 0- = (^3) ✗ (^) 0.1621 MOI = 0.1625 MOI =^ 0-97 M = (^) 0-4863 (^) MOI mass Fe^ " = (^) 0.1625 ✗ 55- = (^) 9.

A solution is prepared by dissolving 3. 5725 g H 3

PO

4 in 500 mL volumetric flask and diluting it to the calibration mark. Calculate the: i) normality of the H 3

PO

4 solution. ii) ppm of H

ion in the solution.

• (^) Based on nuns^ of^ Hydrogen^ (^ H (^) )
• N =^ M.^ n

concentration^ /^ molarity

• Find^ Molarity^ ,^ Find^ Mol g 0 -^ b- (^) L ◦ Find Normality ~ (i)^
• (^) Find (^) molarity I μ :^ no ◦^ t^ equivalent of (^) reacting unit^ leg]^ 3.5725g^ Hzpoy^ 1m01 (^) 1- = Volume of (^) solution (l) (^) I (3) + (^) (30-1)+(16×4) = ◦ '^5 ʰ = 97 -1g

Parts (^) per million^ = 0<0736 M ppm =^ mass^ solute^ Normality = M ✗^ h ✗ (^106) mass sample^ = (^) 0.0736 ✗ (^3) = (^0). 22 N MOI H , Poi, =^ 3. MOI H^ " = 1m01 (^) tlspoy ✗ 3m01 H " ppm = 0.11 (^) mop × Ig^ Ht H ' moi ti × 97.1 (^9) 1m01 1- 1000mg × ' = 0-036 MOI^ = 3 ×0. 1g (^0) -5L = (^0) -11 (^) MOI = (^220) ppm #

Describe the preparation of 1. 00 L of 2. 00 M ammonium hydroxide, NH 4 OH from a concentrated solution that has a purity of 25 % (w/w) and a specific gravity of 0. 957. · (^) molar mass (^) NH, OH MeNc = MdId 14 + 4(1) +^ 16+^1 = 35g/m0) MC:molarity^ of^ concentration^ solution mol of^ solute^ Xc:volume of concentrated^ stock , (^) molarity volume Nd=Molarity^ of^ Dilute solution · (^) concentration NHOH (^) 25%(w/w), gravity^

Vd:Volume^ of^ Dilute solution 2590.9579+1m0 x^ 1000 / X 359 12 1809 mL = 6.8357M McVc = MdXd (6.8357M) Xc^ = (2.00M( /1.002) i. (^) Measured 192.6mL of the^ concentrated NHisH and^ dilute^ it (^) with distilled^ mater V. =^ 0. to the^ mark^ in^ 1000 volumetric^ flash =292.6mL

Methanoic acid (HCOOH) has a density of 1. 22 g/mL. Describe the preparation of 500 mL 0. 70 M HCOOH without weighing the methanoic acid. (^) >^ D=^ MXY Mol HCOOH^ =^ 0-70^ ×^ (^500) =

1000 mass 1-11001-^ = 0 - 35m01 ✗ (21+112)+(16×2) = 16- Volume = (^) mass HCOOH (^) density i. Dissolve 13.2mL^ Of HCOOH in^ distilled = (^) 16. = (^) 13.2mL water in^ a 500mL Volumetric^ flash (^). 1- (^22) Add water (^) up to (^) the calibration mark and shake^ well^.

Hydrochloric acid is usually sold commercially as 37% (w/w) in a bottle of 2.5 L. The bottle label stated its specific gravity at 20°C is 1.19 and its formula weight is 36.46 g/mol. Calculate its molar concentration if the density of water at 20°C is 0.9982 g/mL. HCI molarity = MOI Volume ✗ 1.19 (^) g 1m^ 379 ×^ 0. ✗ I , m , ✗ 1000mL ×

1L (^) 36-469 1 mL 1009 = (^) 12.05M

• (^) HCl from commercial^ reagent (^) , 37 % (^ w/ w^ ) &^ specific (^) gravity = 1.1991mL
• (^) Density of^ HCl^ = specific density ✗^ density of water = I^ -19 ✗ (^) 0.9982 91mL = I (^) -1879 (^) 91mL

(^1). w/w = mass^ of^ solute mass of^ solution ✗ 100% mass of^ = (^379) ✗ 1.^ iptqg^ /me^ ✗^1000 mL solute (^) 100g = (^) 439.52g mop of^ solute^ '^ n = (^) 439.529 (^) = 12.05 MOI 1L solution 36.469 1m Molar concentration^ = 12.05 Mol^ ✗^ 9-^ L = (^) 12.05 M

Explain how to prepare 250 mL 350 ppm Mg 2+ from solid magnesium sulphate Mass of^ = 0.^ ✗ (^) Mgs 04 (^350) ppm =^ mass •^ t^ m^ (9) ✗ 106 Megson mg 250 mL = 0 -0875 ✗ ( 24.30+36-06 +^ (4×16) mass of^ 350 mg (g) = 106 ✗ 25°^ Mt

= 0.4375g = (^) 0. g i (^) _ Dissolve 0.4375 (^) g of Mlg so^ , in^ some^ distilled water^ in 250mL Volumetric^ flask^.^ Then^ add^ distilled water^ up^ to the calibration mark^ and^ Shake^ well to^ make^ the solution homogeneous.

Explain how to prepare 2.0 L solution that is 50.0 ppm in NO 3

• , from solid magnesium nitrate Mg IMO^ }^ ),

(^) Mg 2-

• 2 NO} ' concentrators =^ (^50) ppm =^ 50mg / L MOI (^) Mg CNO^ } ) ≥ ' mass of^ NOJ^ = (^) 50mg / 2 ×^ 2< MOI NO,

= 2 = 100mg = (^100) = (^) 0.1g 1000 MOI Mg (NO^ }) , = ÷ ✗^ (1.613×10-3) Molar mass^ NO,

= 14+3116)^ =^629 / n° = (^) g. 065 ✗^10 _ " Mot (^) Noi =^ ◦^ -^ molar mqgg Mg^ (Mp3 ),^ =^ 24+2441+^ ) 6 2g 1m = 1- 613 ✗^ 10-3^ may = (^148) 91m Mass (^) mgl.NO } ), = 8- 065 ✗^ 10-4^ MOI^ ✗^1489 / mop Magnesium Nitrate =^ My^ CNO (^) ,) (^) , = 0 - 1194g i. (^) weigh 0-1194^9 Of^ Mg^ (^ Nos),^ , dissolve in^ water and transfer^ Into^ 2.0L^ volumetric^ flask and^ dilute it to^ the^ mark^.