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A- preparation of approximately (0.1N) HCL 1. Calculate the normality of the concentrated HCL. Sp.g Percentage 1000 N= القانون الخاص بتحضير المحاليل السائلة equival ent wei ght equivalent weight = Molecular weight / no. of Hydrogen Sp.gr = 1.19 gm/cm³ Percentage = 37 % (35.5 + 1) = 36.5 eq.wt (HCl) = - H=1 CI=35.5 1.19 (37/100) ×1000 N= =12.0630 36.5 2-To prepare (500ML) of 0.1N HCI N¡ × Vị conc.HCI N2 x V2 dil.HCl 12.0630 × Vị = 500 × 0.1 VI 4.1449 mL %3D

A- preparation of approximately (0.1N) HCL 1. Calculate the normality of the concentrated HCL. Sp.g Percentage 1000 N= القانون الخاص بتحضير المحاليل السائلة equival ent wei ght equivalent weight = Molecular weight / no. of Hydrogen Sp.gr = 1.19 gm/cm³ Percentage = 37 % (35.5 + 1) = 36.5 eq.wt (HCl) = - H=1 CI=35.5 1.19 (37/100) ×1000 N= =12.0630 36.5 2-To prepare (500ML) of 0.1N HCI N¡ × Vị conc.HCI N2 x V2 dil.HCl 12.0630 × Vị = 500 × 0.1 VI 4.1449 mL %3D

Principles of Modern Chemistry
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter11: Solutions
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Problem 1P: A patient has a “cholesterol count” of 214. Like manyblood-chemistry measurements,this result is...
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i li. In. 3 ۱ ۱:۳۸ 88 EXPERIMENT A- preparation of approximately (0.1N) HCL B- preparation of (0.1N) Sodium carbonate Na2CO3 The aim of the experimental:- -help the student know how to prepare solution. -to know the techniques involves of standardization solution. - determine the concentration of an unknown solution( HCI ). A- preparation of approximately (0.1N) HCL 1. Calculate the normality of the concentrated HCL. Sp.g • Percentage • 1000 N= القانون الخاص بتحضير المحاليل السائلة equival ent wei ght equivalent weight = Molecular weight / no. of Hydrogen Sp.gr = 1.19 gm/em Percentage = 37 % (35.5 + 1) = 36.5 eq.wt (HCl) = H=1 Cl=35.5 N- 1.19- (37/100) × 1000 =12.0630 36.5 2-To prepare (500ML) of 0.1N HCI N¡ × Vị N2 × V2 conc.HOCI dil.HCl 12.0630 × Vị = 500 × 0.1 Vj= 4.1449 mL B- preparation of (0.1N) Sodium carbonate Na2cO3 Weight 1000 القانون الخاص لتحضير المواد الصلبة equivalent weight Volume(mL ) M. mass (23x2)+12+(16x3) 106 aCO;) = total charge for positive ion = 53 2 II >
Transcribed Image Text:i li. In. 3 ۱ ۱:۳۸ 88 EXPERIMENT A- preparation of approximately (0.1N) HCL B- preparation of (0.1N) Sodium carbonate Na2CO3 The aim of the experimental:- -help the student know how to prepare solution. -to know the techniques involves of standardization solution. - determine the concentration of an unknown solution( HCI ). A- preparation of approximately (0.1N) HCL 1. Calculate the normality of the concentrated HCL. Sp.g • Percentage • 1000 N= القانون الخاص بتحضير المحاليل السائلة equival ent wei ght equivalent weight = Molecular weight / no. of Hydrogen Sp.gr = 1.19 gm/em Percentage = 37 % (35.5 + 1) = 36.5 eq.wt (HCl) = H=1 Cl=35.5 N- 1.19- (37/100) × 1000 =12.0630 36.5 2-To prepare (500ML) of 0.1N HCI N¡ × Vị N2 × V2 conc.HOCI dil.HCl 12.0630 × Vị = 500 × 0.1 Vj= 4.1449 mL B- preparation of (0.1N) Sodium carbonate Na2cO3 Weight 1000 القانون الخاص لتحضير المواد الصلبة equivalent weight Volume(mL ) M. mass (23x2)+12+(16x3) 106 aCO;) = total charge for positive ion = 53 2 II >
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