Đáp án đúng: C

Phương pháp giải:

Este có CTTQ:  CnHnOn-6   ⟹ n chẵn, n > 6  ⟹ n = {8, 10, …}Mà: Meste < 160  ⟹ n = 8  ⟹ C8H8O2⟹ \(X\left\{ \begin{array}{l}HCOO - C{H_2} - {C_6}{H_5}:{a^{mol}}\\{C_6}{H_5}COOC{H_3}:{b^{mol}}\\HCOO{C_6}{H_4} - C{H_3}\left( {x3} \right):{c^{mol}}\\C{H_3}COO{C_6}{H_5}:{d^{mol}}\end{array} \right. + KOH \to \left\{ \begin{array}{l}F\left\{ \begin{array}{l}HCOOK\\C{H_3} - {C_6}{H_4} - OK\\{C_6}{H_5} - OK\\C{H_3}COOK\\{C_6}{H_5}COOK:{b^{mol}}\end{array} \right.\\{H_2}O:c + d\\Z\left\{ \begin{array}{l}{C_6}{H_5} - C{H_2} - OH:{a^{mol}}\\C{H_3}OH:{b^{mol}}\end{array} \right. \to \left\{ \begin{array}{l}{C_6}{H_5} - CHO\\HCHO\end{array} \right.\end{array} \right.\)Từ mZ và nAg  ⟹ a , bBTKL: mX + mKOH = mF + mZ + mH2OMà: nH2O = nX - nZ nKOH = nX ⟹ mF ⟹ %mC6H5COOK Giải chi tiết:Este có CTTQ:  CnHnOn-6   ⟹ n chẵn, n > 6  ⟹ n = {8, 10, …}Mà: Meste < 160  ⟹ n = 8  ⟹ C8H8O2⟹ \(X\left\{ \begin{array}{l}HCOO - C{H_2} - {C_6}{H_5}:{a^{mol}}\\{C_6}{H_5}COOC{H_3}:{b^{mol}}\\HCOO{C_6}{H_4} - C{H_3}\left( {x3} \right):{c^{mol}}\\C{H_3}COO{C_6}{H_5}:{d^{mol}}\end{array} \right. + KOH \to \left\{ \begin{array}{l}F\left\{ \begin{array}{l}HCOOK\\C{H_3} - {C_6}{H_4} - OK\\{C_6}{H_5} - OK\\C{H_3}COOK\\{C_6}{H_5}COOK:{b^{mol}}\end{array} \right.\\{H_2}O:c + d\\Z\left\{ \begin{array}{l}{C_6}{H_5} - C{H_2} - OH:{a^{mol}}\\C{H_3}OH:{b^{mol}}\end{array} \right. \to \left\{ \begin{array}{l}{C_6}{H_5} - CHO\\HCHO\end{array} \right.\end{array} \right.\)⟹ 108.a + 32b = 1,72 và: 2a + 4b = nAg = 10,8/108 = 0,1⟹ a = 0,01;  b = 0,02.BTKL: mX + mKOH = mF + mZ + mH2OMà: nH2O = nX - nZ = 0,05 - 0,01 - 0,02 = 0,02nKOH = nX = 0,05⟹ 0,05.136 + 0,05.56 = mF + 1,72 + 0,02.18  ⟹ mF = 7,52 (g)⟹ %mC6H5COOK = 0,02.160/7,52 = 0,4255 = 42,55%