JEE Advance - Chemistry (2021 - Paper 2 Online - No. 9)
Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).
(Use Molar masses (in g mol$$-$$1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively).
The value of x is _________.
(Use Molar masses (in g mol$$-$$1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively).
The value of x is _________.
Answer
3.57
Explanation
Mass of organic salt produced (aniline) = 1.29 g
Molar mass of organic salt (aniline)
= 12 × 6 + 1 × 8 + 14 × 1 + 35 × 1
= 72 + 8 + 14 + 35
= 129 g/mol
$$Moles\ of\ organic\ salt=\frac{Mass\ of\ organic\ salt}{Molar\ mass} $$
$$ =\frac{1.29}{129} =0.01\ mol $$
From reaction 1 moles of salt is produced from 3 mole of Sn. So, 0.01 mole of organic salt is produced by 0.03 mole Sn. Atomic mass of Sn = 119 g mol−1
Mass of Sn = x = mole of Sn × Molar mass
x = 0.03 × 119 $$ \Rightarrow $$ x = 3.57 g
The value of x is 3.57
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