A reaction of 0.1 mole of Benzylamine with bromomethane gave 23g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n x 10?¹, when n=_____. (Round off to the nearest integer).
[Given: Atomic masses: C=12.0u, H:1.0u, N:14.0u, Br:80.0u]

8 Views|Posted 5 months ago
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5 months ago

Benzyl amine (0.1 mole) reacts with 3 equivalents of CH? Br to form Benzyl trimethyl ammonium bromide.

Given 23g of Benzyl trimethyl ammonium bromide (molar mass 230 g/mol ), which is 0.1 mol.

Therefore, moles of CH? Br = 0.3 = 3 x 10? ¹. The value of n is 3.

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E_cell = E? _cell - (0.059/2)log { [Cu²? ]/ [Ag? ]²}
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E? _cell = 0.398

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E? = 0.398 - (0.059/2) x 4 V

E? = 0.28 V
E? = 28 x 10? ² V

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= 1.05 - (0.059 / 2) log ( [10? ³] / [10? ³]²)
= 1.05 - (0.059 × 3) / 2 = 1.05 - 0.0885 = 0.9615 volt
There is a misprint in the question. The E? cell is incorrectly given as 10.5 V. This should have been 1.05 volt.

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