First, we shall calculate the total number of moles present in the final solution.
Number of moles in 0.50 m NaCl = molarity * volume = 0.50 * 3.0 = 1.5 moles.
Number of moles in 0.2777m NaCl = molarity * volume = 0.2777 * 9.0 = 0.24993 moles
Total number of moles = 1.5 + 0.24993 = 1.74993 moles
Second, we shall calculate the total volume of the final solution.
Total volume = 3 + 9 = 12 litres.
The molarity = total number of moles / total volume = 1.74993 / 12 = 0.1458 m
In nitration u add two concentration acids HNO3 and H2SO4 both are strong acids. always the acid base neutralization reactions are fastest. even though NH2 group in aniline is a weak base, its neutralization with a very strong acid is still faster than nitration. hence if u carry the acid base reaction u would end up with an NH3+ group. hence the meta substitution can't be rule out here and u would get all products ortho,meta and para...
The answer is 59.01 g
If a 100-gram sample of NaCl consists of 39.34 g of sodium, a 150-gram sample of NaCl will consist x g of sodium:
100 g : 39.34 g = 150 g : x
x = 39.34 g * 150 g : 100 g
x = 59.01 g