5mL of NHCl, 20 mL of N/2 H_{2}SO_{4} and 30mL of N/3 HNO_{3} are mixed together and volume made one litre. The normality of the resulting solution is:

(a) N/5

(b) N/10

(c) N/20

(d) N/40

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Equivalent Weight & Normality

(d) M eq. of HCl = 5 x1 =5;

M eq. of H_{2}SO_{4} = 20 x (1/2) = 10;

M eq.of HNO_{3} = 30 x (1/3) =10;

Thus, total M eq. of acid = 5+10+10=25

Total volume = 1000mL

Also M eq. = N x V,

N = 25/1000=1/40

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When a metal is burnt, its mass is increased by 24 percent. The equivalent mass of the metal will be:

(a) 25

(b) 24

(c) 33.3

(d) 76

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Equivalent Weight & Normality

(c) Equivalent of metal = Equivalent of oxide

100/E = 24/8

E = 33.3

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1g of pure calcium carbonate was found to require 50 mL of dilute HCl for complete reactions. The strength of the HCl solution is given by:

(a) 4N

(b) 2N

(c) 0.4N

(d) 0.2N

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Equivalent Weight & Normality

(c) M eq. of HCl = M eq. of CaCO_{3};

therefore N x 50 = 1/50*1000 or N = 0.4

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The equivalent mass of H_{3}PO_{4} in the following reaction is,

H_{3}PO_{4 }+ Ca(OH)_{2} $\to $CaHPO_{4} + 2H_{2}O

(a) 98

(b) 49

(c) 32.66

(d) 40

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Equivalent Weight & Normality

(b) Equivalent mass = molar mass / basicity

M/2=98/2=49;

Basicity =2; Only two H are replaced

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1.520g of the hydroxide of a metal on ignition gave 0.995 g of oxide. The equivalent mass of metal is:

(a) 1.520

(b) 0.995

(c) 19.00

(d) 9.00

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Equivalent Weight & Normality

(d) M eq. of oxide = M eq. of hydroxide;

Thus, 0.995/E+8 = 1.520/E+17

E = 9

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0.5 g of fuming H_{2}SO_{4} (oleum) is diluted with water. This solution is completely neutralised by 26.7 mL of 0.4 N NaOH. The percentage of free SO_{3} in the sample is:

(a) 30.6%

(b) 40.6%

(c) 20.6%

(d) 50%

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Equivalent Weight & Normality

(c) Meq. of H_{2}SO_{4} + Meq.Of SO_{3} = Meq. of NaOH

(0.5-a)/49*1000 +a/40*1000 = 26.7 x 0.4

a = 0.103

% of SO_{3} = 0.103/0.5*100 = 20.6%

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One g of a mixture of Na_{2}CO_{3} and NaHCO_{3} consumes y equivalent of HCl for complete neutralisation. One g of the mixture is strongly heated, then cooled and the residue treated with HCl. How many equivalent of HCl would be required for complete neutralisation?

(a) 2y equivalent

(b) y equivalent

(c) 3y/4 equivalent

(d) 3y/2 equivalent

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Equivalent Weight & Normality

(b) 2NaHCO_{3} $\stackrel{\u2206}{\to}$Na_{2}CO_{3} +H_{2}O + CO2

Na_{2}CO_{3} $\stackrel{\u2206}{\to}$Na_{2}CO_{3}

The no. of equivalent of NaHCO3 = No. of equivalent of NaHCO_{3} formed. Thus, same equivalent of HCl will be used.

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The chloride of a metal contains 71% chlorine by mass and the vapour density of it is 50. The atomic mass of the metal will be:

(a) 29

(b) 58

(c) 35.5

(d) 71

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Equivalent Weight & Normality

(a) Molar mass of metal chloride = 50x2=100;

Let metal chloride be MCl_{n} then

Equivalent of metal = Equivalent of chloride, or 29/E = 71/35.5

E=29/2

Now a+35.5n =100

or n.E+35.5n = 100

n=2

therefore a = 2 x E= 2x(29/2) = 29.3

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The equivalent mass of Zn(OH)_{2} in the following reaction is equal to its,

Zn(OH)_{2} + HNO_{3} $\to $Zn(OH)(NO_{3}) + H_{2}O :

(a) Formula mass/1

(b) Formula mass/2

(c) 2 x formula mass

(d) 3 x formula mass

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Equivalent Weight & Normality

(a) Equivalent mass Zn(OH)_{2} = molar mass/acidity = M/1

Acidity of Zn(OH)_{2} = 1; only one OH is replaced.

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What will be the normality of a solution obtained by mixing 0.45 N and 0.60 N NaOH in the ratio 2:1 by volume?

(a) 0.4 N

(b) 0.5 N

(c) 1.05 N

(d) 0.15 N

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Equivalent Weight & Normality

(b) M eq.Of NaOH = 0.45 x 2V + 0.6xV

Total volume = 3V

N x 3V = 0.45 x 2V + 0.6V

N = 0.5

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