kimyo (eritma)

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Yangilangan May 2026

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@vokhidov_bekzod @vokhidov_sardor 1 eritma 1. ma’lum temperaturada naoh ning eruvchanligi 20 ga teng. 327 g suvda qancha (g) na eritilsa to‘yingan eritma hosil bo‘ladi? a) 28,75 b) 40,25 c) 23 d) 34,5 yechim: na + h2o = naoh + ½h2 23x --18x ---- 40x 100 g (h2o) ---------------- 20 g (naoh) 327 - 18x ---------------- 40x 327 – 18 x = 200x x = 1,5 23x = 34,5 g (na) javob: d) 34,5 2. 20 °c dagi eruvchanligi 42 g bo‘lgan tuzning 568 g to‘yingan eritmasi 40 °c gacha isitilganda yana qancha (g) tuz erib, to‘yingan eritma hosil qiladi? s(40°c) = 57 a) 33 b) 66 c) 45 d) 60 yechim: 1 – usul: (20 ℃) 100 ---------- 42 ------------ 142 400=x-------168=x--------- 568 (40 ℃) 100 ---------- 57 400 ---------- x = 228 m(+tuz) = 228 -168 = 60 g 2 – usul: (20 ℃) 100 ---------- 42 ------------ 142 …

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v = 1,2 • 500 = 600 g 100 % ------------- 600 100 % ------------- x = 2000 40 % ------------ x = 240 12 % ----------------- 240 m(+h2o) = 2000 – 600 = 1400 2 – usul: m = ρ • v = 1,2 • 500 = 600 g 40 % 12 -------------- 600 g 12 % 0 % 28 -------------- x = 1400 g javob: c) 1400 7. 720 g (p = 1,2 g/ml) 0,3m li sulfat kislota eritmasi bilan 200 ml 0,6m li x eritmasi to‘liq reaksiyaga kirishdi. x ni toping. a) koh b) ca(oh)2 c) nh3 d) al(oh)3 yechim: v = 720 1,2 = 600 ml n(h2so4) = 0,3 • 0,6 = 0,18 mol n(x) = 0,6 • 0,2 = 0,12 mol 0,12 — 0,18 x + h2so4 2al(oh)3 + 3h2so4 = al2(so4)3 + 6h2o 2 3 javob: d) al(oh)3 8. 8 g naoh dan foydalanib …

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°c 50 °c 100 ---------- 20 ---------- 120 100 ------- 60 200=x ------ 40=x -------- 240 200 ------- x=120 120 – 40 = 80 g (qo‘shish kerak) javob: a) to‘yinmagan eritmaga aylanadi va 80 g eruvchi qo‘shilsa to‘yinadi 12. 300 ml 15 % li naoh eritmasiga (p = 1,2 g/ml) 13,8 g na bo‘lakchasi solindi. olingan eritmadagi ishqorning massasini (g) hisoblang. a) 80 b) 84 c) 82 d) 78 yechim: 13,8 ---------- x = 24 m = 300 • 1,2 = 360 g na + h2o = naoh + ½h2 360 ------------ 100 % 23 ------------- 40 54 = x --------- 15 % m(naoh) = 24 + 54 = 78 g javob: d) 78 13. kumush nitrat eritmasiga ikki marta ko‘p hajmdagi kalsiy xlorid eritmasi qo‘shildi. hosil bo‘lgan eritmada kalsiy va xlorid ionlari 3:2 mol nisbatda bo‘lsa, dastlabki eritmalarning molar konsentratsiyalari qanday nisbatda (berilgan tartibda) bo‘lgan? a) 3:5 b) 8:3 …

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-- 10 % n(o) = x • na m(h2o) = 504x (g) ----------- 90 % n(o) = 28x • na mu = 560x nu(o) = (x + 28x) • na = 5,8 • na x = 0,2 mu = 560x = 560 • 0,2 = 112 g javob: b) 112 17. 40 g 30 % li natriy gidroksid eritmasida qanday massadagi (g) fosfat angidrid eritilganda ekvimolar nisbatdagi nordon tuzlar aralashmasi hosil bo‘ladi? a) 56,8 b) 14,2 c) 28,4 d) 42,6 yechim: m(naoh) = 40 • 0,3 = 12 g 12 -------- x = 14,2 g 3naoh + p2o5 = nah2po4 + na2hpo4 120 ------- 142 javob: b) 14,2 18. 40 g natriy gidroksid eritmasida 14,2 g fosfat angidrid eritilganda ekvimolar nisbatdagi nordon tuzlar aralashmasi hosil bo‘ldi. ishqor eritmasining konsentratsiyasini (%) hisoblang. a) 10 b) 40 c) 30 d) 20 yechim: 12 = x ---- 14,2 g 3naoh + p2o5 = …

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32,1 g uch valentli metall asosi bilan 900 ml 0,5 mol/l sulfat kislota eritmasi to‘liq reaksiyaga kirishdi. metallni aniqlang. a) al b) fe c) mn d) cr yechim: 0,3 = x ------- 0,45 mol 2me(oh)3 + 3h2so4 = me2(so4)3 + 6h2o 2 ------------ 3 m(me) = 32,1 0,3 – (17 • 3) = 56 (fe) javob: b) fe 23. 250 g 20 % li natriy gidroksid eritmasida 10 g naoh eritilganda hosil bo‘lgan eritmaning konsentratsiyasini (mol/kg) aniqlang. (molyal konsentratsiya – 1 kg erituvchiga to‘g‘ri keladigan erigan moddaning mol miqdori.) a) 5,25 b) 7,5 c) 4,5 d) 6,25 yechim: 100 % -------------- 250 g 20 % --------------- x = 50 g (naoh) 80 % --------------- x = 200 g (0,2 kg) (h2o) n(naoh) = 50+10 40 = 1,5 mol cm = 1,5 0,2 = 7,5 mol/kg javob: b) 7,5 24. 250 g 20 % li natriy gidroksid eritmasida necha gramm naoh …

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@vokhidov_bekzod @vokhidov_sardor 1 eritma 1. ma’lum temperaturada naoh ning eruvchanligi 20 ga teng. 327 g suvda qancha (g) na eritilsa to‘yingan eritma hosil bo‘ladi? a) 28,75 b) 40,25 c) 23 d) 34,5 yechim: na + h2o = naoh + ½h2 23x --18x ---- 40x 100 g (h2o) ---------------- 20 g (naoh) 327 - 18x ---------------- 40x 327 – 18 x = 200x x = 1,5 23x = 34,5 g (na) javob: d) 34,5 2. 20 °c dagi eruvchanligi 42 g bo‘lgan tuzning 568 g to‘yingan eritmasi 40 °c gacha isitilganda yana qancha (g) tuz erib, to‘yingan eritma hosil qiladi? s(40°c) = 57 a) 33 b) 66 c) 45 d) 60 yechim: 1 – usul: (20 ℃) 100 ---------- 42 ------------ …

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