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Equivalent mass and n-factor for acids, bases and salts
JEE Main level
Single correct option
Difficulty 2 of 3
eq123
How many equivalents are present in 14.3 g of
Na
X
2
CO
X
3
⋅
10
H
X
2
O
(molar mass 286 g/mol) when it is titrated with excess HCl to
CO
X
2
?
A
0.10
B
0.05
C
0.20
D
0.025
Answer: A
Moles = 14.3/286 = 0.0500. Each mole of carbonate takes 2
H
X
+
, so n = 2 and equivalents = moles
×
n = 0.0500
×
2 = 0.10.
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