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]
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]
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Ecell = 0 –
= – 0.03 (0.3)
= – 0.009
= – 9 * 10–3 V
x = 9
H2 (g) + Cu2+ (aq) 2H+ (aq) + Cu (s)
0.31 = 0.34 -
x = 7
H2 (g) + Cu2+ (aq) 2H+ (aq) + Cu (s)
0.31 = 0.34 -
x = 7
E_cell = E? _cell - (0.059/2)log { [Cu²? ]/ [Ag? ]²}
E? = E? _cell - (0.059/2)log [0.1/ (0.01)²] = 0.3095
E? _cell = 0.3095 + 0.0885 V
E? _cell = 0.398
Again, E? = E? _cell - (0.059/2)log [10? ²/ (10? ³)²]
E? = 0.398 - (0.059/2) x 4 V
E? = 0.28 V
E? = 28 x 10? ² V
Ecell = 1.05 - (0.059 / 2) log ( [Ni²? ] / [Ag? ]²)
= 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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Chemistry NCERT Exemplar Solutions Class 11th Chapter Three 2025
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