Answer on Other Chemistry Question for somnath chatterjee
2)how many gms of a dibasic acid (mol wt 200) should be present in 100ml of its aqueous solution to give 0.1(N) strength?
1 L of 0.5 M HCl solution contain n(HCl) = c(HCl)*V(1 M HCl solution) = 1 L * 0.5 M = 0.5 mol of HCl.
The volume of 6 M HCl solution needed to prepare the 1 M HCl solution is
V(6 M HCl solution) = n(HCl)*c(HCl) = 0.5 mol / 6 M = 0.083 L = 83 mL
The molar mass of the acid is 200 g/mol
The volume of the acid solution is 0.1 L
The normality of the acid solution is 0.1 N
The normality of a solution is the molar concentration divided by an equivalence factor: N(acid) = c(acid) / f(eq), c(acid) = N(acid) * f(eq)
The equivalence factor of a diprotic acid is 0.5.
The concentration of the dibasic acid solution is
c(acid) = n(acid)/V(acid)
n(acid) = c(acid) * V(acid)
n(acid) = m(acid)/M(acid)
Thus, the mass of acid will be
m(acid) = n(acid) * M(acid) = c(acid) * V(acid) *M(acid) = N(acid) * f(eq) * V(acid) * M(acid) = 0.1 N * 0.5 * 0.1 L * 200 g/mol = 1 g
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