Answer:
e. 0.0822 M
Explanation:
Considering:-
[tex]Normality=Molarity\times {n-factor}[/tex]
So, Given that:-
Molarity of [tex]Na_3PO_4[/tex] = 1.28 M
n-factor of [tex]Na_3PO_4[/tex] = 3
So,
Normality of [tex]Na_3PO_4[/tex] = 3*1.28 N = 3.84 N
Considering:-
At equivalence point
Gram equivalents of [tex]Cd^{2+}[/tex] = Gram equivalents of [tex]Na_3PO_4[/tex]
So,
[tex]Normality_{Cd^{2+}}\times Volume_{Cd^{2+}}=Normality_{Na_3PO_4}\times Volume_{Na_3PO_4}[/tex]
Given that:
[tex]Normality_{Na_3PO_4}=3.84\ N[/tex]
[tex]Volume_{Na_3PO_4}=77.1\ mL[/tex]
[tex]Volume_{Cd^{2+}}=1.80\ L=1800\ mL[/tex]
So,
[tex]Normality_{Cd^{2+}}\times 1800=3.84\times 77.1[/tex]
[tex]Normality_{Cd^{2+}}=\frac{296.064}{1800}=0.16448\ N[/tex]
Also,
n-factor of [tex]Cd^{2+}[/tex] = 2
So, [tex]Molarity=\frac{Normality}{n-factor}=\frac{0.16448}{2}=0.0822\ M[/tex]
Hence, e is the answer.