Cty TIMHH MTV DVVH Khang Vigt Ca'm nang 6n luygn thi dgi hgc 18 chuygn dg H6a hgc - Nguygn Van htai ''^''[ mstm^ ' • V ^ y : m = 127.0,1 + 74,5.0,2 = 27,60 gam - > Dap an B ' ' ' ' ' r = - ^ i man ^ Lin< 1/: Trong bai cac em de quen KCl h'nh khoi lugng muol va chor, nham phuong an D! ^ xet: Vi d\ 6: Cho 0,02 mol amino axit X tac dung vira dii voi 10ml dung dich HC| A H2NC2H3(COOH)2 C (H2N)2C3H5COOH • D H2NC4H8COOH 'f^^'Ky: ^ Laigidi: gam X can 60ml dung djch HCl I M Cho 7,66 gam X tac dung vira du vol dung dich KOH, thu dugc dung dich Y Co can Y thu dugc m gam muoi "^"^ khan Gia tri cua m la A 11,31 B 11,46 - Ta c6: nHci: nx = 1:1 —> Trong X c6 chiia nhom - N H " N a O H : nx = 2:1 H2N-R-COOH Trong X chua nhom - C O O H + HCl 0,02 • "KOH > ClHaN-R-COOH ' + '-i' R = 56 (C4H8) = 153,5 Dap an D vai 100ml dung dich NaOH 2M Cong thuc cau tgo cua X la: A H2N-C3H6-COOH Laigidi: mol; nwaOH = 0,02 Nhan xet: nna: nx = 1:1 nNaOH : nx ' •;„;••.••,: •• 10.0,06 , ^ n o ( X ) = — ^ = , mol ^ R + 142,5 = ;f ' J AMti'Dr :y r / * ncooH V f ' , ^ = 0/1 mol , , » B Valin C Lysin D Alanin '• Lai gidi: |j D 0,55 Phuong an C: Lysin c6 cong thuc cau tao (NH2)2C5H9COOH W^J^-tHD dung dich CO moi truong b a z a l a m quy tim chuyen xanh HCl 2M, thu dugc dung djch X Cho NaOH du vao X Sau cac phan ung xay hoan toan, somol NaOH da phan ling la iiCJg -i^n' Trong phan tir lysin c6 nhom chuc amin va nhom chuc axit cacboxyl -> = 183,5 -> R = 41 (C3H5) ^ Dap an C C.0,65 ' ' ^ , lam doi mau qui tim > ClH3NR(COOH)2 0,1 B.0,50 10 Loai A, B, D vi dung dich cac chat c6 moi truong trung tinh va khong V i du 8: Cho 0,15 mol H2NC3H5(COOH)2 (axit glutamic) vao 175ml dung dich A 0,70 80/16 10: Dung dich nao sau day lam chuyen mau qui tim xanh A Glyxin = 2:1 -> X chua nhom - C O O H - ,?* iv- 0,1 no ! -> Dap an B —> X chua nhom -NH2 Ggi cong thuc cua X la H2N-R(COOH)2 Mol: Matkhac: no(x)=2ncooH ^ H mol H2NR(COOH)2 + H C l 80 Tang giam khoi lugng: m = mx + 0,1(39-1) = 7,66 + 0,1.38 = 11,46 gam., H D.(H2N)2C3H5-COOH C.H2N-C3H5(COOH)2 n H c i = 0,01 "^o Nhqn thay: Khi tac dung voi KOH, - C O O H se chuyen - C O O K nen: B H2N-C2H3(COOH)2 '•••••"•r T - ' i - i u - i T u ti 1? khoi luong:—— = — -» = = — • ^ mN 21 nN 21/14 I Cac em can luu y: n N = n N H = n H c l - > n N ( x ) = 0,06 mol I M , thu dugc 18,35 gam muoi khan Mat khac, 0,1 mol X tac dung vira dii ^ xiw^ N/iflM Aret: I De thay: nf,jH9 = " H C l = 0/06 mol Vi dv 7: Cho X la mot amino axit Cho 0,1 mol X tac dung vira du 100 ml HCl I D 9,12 =0/1 mol- ^a02 07 -> ClHaN-R-COOH = C.9,85 Laigidi: G