$\rm Ag^{+}+SCN^- \to AgSCN↓$

$\rm 10 \times C_N = 0.0467 \times 11.75$

`=>`$\rm C_N=C_M=0.05487(N)$

`=>`$n_{Ag}=0.5487 \times 10^{-3}(mol)$

Trong $250(mL)$ sẽ là : $n_{Ag}=1.372 \times 10^{-2}(mol)$

`=>`$\%m_{Ag}=\dfrac{1.3718 \times 10^{-2} \times 108}{1.7450} \times 100\%=84.91\%$

Đáp án:

Tại thời điểm đương lượng:

$\rm N_{Ag^+}=\dfrac{V_{NH_4SCN}×N_{NH_4SCN}}{V_{Ag^+}}=\dfrac{11,75×0,0467}{10}=0,0549\ (N)$

$\rm Ag^++SCN^-\to AgSCN$

$\rm C_{Ag^+}=\dfrac{N_{Ag^+}}{\gamma}=\dfrac{0,0549}{1}=0,0549\ (M)$

$\rm n_{Ag^+}=0,0549×10×10^{-3}=5,49×10^{-4}\ (mol)$

Trong $\rm 250$ mL có:

$\rm n_{Ag^+}=5,49×10^{-4}×25=1,3725×10^{-2}\ (mol)$

$\rm\%m_{Ag^+}=\dfrac{1,3725×10^{-2}×108}{1,745}×100\%=84,95\%$