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
Introductory Chemistry: An Active Learning Approach
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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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