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The equivalent mass of H3PO4 in the following reaction is,
H3PO4 + Ca(OH)2 CaHPO4 + 2H2O
(b) Equivalent mass = molar mass / basicity
M/2=98/2=49;
Basicity =2; Only two H are replaced
One g of a mixture of Na2CO3 and NaHCO3 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?
(b) 2NaHCO3 Na2CO3 +H2O + CO2
Na2CO3 Na2CO3
The no. of equivalent of NaHCO3 = No. of equivalent of NaHCO3 formed. Thus, same equivalent of HCl will be used.
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) Molar mass of metal chloride = 50x2=100;
Let metal chloride be MCln 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
The equivalent mass of Zn(OH)2 in the following reaction is equal to its,
Zn(OH)2 + HNO3 Zn(OH)(NO3) + H2O :
(a) Equivalent mass Zn(OH)2 = molar mass/acidity = M/1
Acidity of Zn(OH)2 = 1; only one OH is replaced.
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?
(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
0.7 g of Na2CO3.xH2O were dissolved in water and the volume was made to 100mL, 20mL of this solution required 19.8 mL of N/10 HCl for complete neutralisation. The value of x is:
(c) M eq. of Na2CO3.xH2O in 20 mL = 19.8x1/10
M eq. of Na2CO3.xH2O in 100 mL = 19.8x1/10x5
w/E*1000=19.8x1/10x5
or 0.7/M/2*1000=19.8/2
M=141.41
23x2+12+3x16+18x= 141.41
x=2
A metal oxide is reduced by heating it in a stream of hydrogen. It is found that after complete reduction, 3.15 g of the oxide have yielded 1.05g of the metal. We may deduce that:
(c) Equivalent of metal = Equivalent of oxygen;
Thus, 1.05/E = 3.15-1.05/8
E=4
An oxide of metal has 20% oxygen, the equivalent mass of oxide is:
(b) Equivalent of metal oxide = Equivalent of oxygen
E=40
How much water is to be added to dilute 10 mL of 10N HCl to make it decinormal?
(a) M eq. of conc.HCl = M eq. of dil. HCl;
10x10=Vx1/10
V=1000mL
Thus, 990mL of water should be added to 10mL if conc. HCl to get decinormal solution.
W1 g of an element combines with oxygen forming W2 g of its oxide. The equivalent mass of the element is:
(b) Equivalent of element = Equivalent of oxygen or W1 /E1=W2-W1/8
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