JEE MAIN - Chemistry (2020 - 4th September Evening Slot - No. 7)
A 100 mL solution was made by adding 1.43 g
of Na2CO3.xH2O. The normality of the solution
is 0.1 N. The value of x is _____.
(The atomic mass of Na is 23 g/mol)
(The atomic mass of Na is 23 g/mol)
Answer
10
Explanation
Molar mass of Na2CO3.xH2O
= 23 × 2 + 12 + 48 + 18x
= 46 + 12 + 48 + 18x
= (106 + 18x)
As nfactor in dissolution will be determined from net cationic or anionic charge; which is 2
Eq wt = $${M \over 2}$$ = (53 + 9x)
volume = 100 ml = 0.1 Litre
Normality =
$$ \Rightarrow $$ 0.1= $${{{{1.43} \over {53 + 9x}}} \over {0.1}}$$
$$ \Rightarrow $$ 53 + 9x = 143
$$ \Rightarrow $$ 9x = 90
$$ \Rightarrow $$ x = 10
= 23 × 2 + 12 + 48 + 18x
= 46 + 12 + 48 + 18x
= (106 + 18x)
As nfactor in dissolution will be determined from net cationic or anionic charge; which is 2
Eq wt = $${M \over 2}$$ = (53 + 9x)
volume = 100 ml = 0.1 Litre
Normality =
No. of equivalents of solute
Volume of solution (in L)
$$ \Rightarrow $$ 0.1= $${{{{1.43} \over {53 + 9x}}} \over {0.1}}$$
$$ \Rightarrow $$ 53 + 9x = 143
$$ \Rightarrow $$ 9x = 90
$$ \Rightarrow $$ x = 10
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