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Noma’lum konsentrasiyali va nomalum massadagi eritmalar
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Noma’lum konsentrasiyali va nomalum massadagi eritmalar
1. 150 g X % li NaCl eritmasi elektroliz qilinganda 11,2 litr gazlar ajraldi. Anodda ajralib chiqqan gazlar 1:1 mol nisbatda bo’lsa, hosil bo’lgan eritma necha foizli bo’ladi? A) 5,75 B) 5,33 C) 6,75 D) 7,45 Yechimi: . 1. 1 2NaCl+2H2O=H2+Cl2+2NaOH . 2. 1
Jami gaz 5 mol 5x=11,2 x=2,24 l 2,24/22,5=0,1mol Cl2(n) 0,1 mol O2(n) 0,1 mol Demak:
2NaCl+2H2O=H2+Cl2+2NaOH
Reaksiya bo'yicha: H2(m) 0,1×2=0,2 g Cl2(m) 0,1×71=7,1 g 2NaOH(m) 0,1×80=8 g 2H2O=2H2+O2 2H2O(m) 36×0,1=3,6 g . 8 g
. 150-0,2-7,1-3,6 =5,75%
J: A
2. X g 40% li CaBr2 eritmasini to’yintirish uchun 20 g CaBr2 tuzidan qo’shish kerak. Tuzning ushbu haroratda eruvchanlik koeffitsiyenti 80 ga teng. Boshlang’ich eritmada qancha miqdorda (g) tuz qolgan? A)250 B)150 C)100 D)300 Yechimi:
180. 80
x+20. 0,4x+20 x=250 g dastlabki eritma massasi CaBr2(m) 250×0,4=100 g J: C
3. 4 g tuz 42g x % li eritmaga qo'shilganda to'yingan eritmaga aylanadi . Shu tuzning eruvchanlik koeffitsiyenti 15 ga teng bo'lsa , x qiymatini aniqlang.
A) 4,76 B) 2,26 C) 13,04 D)14,3 Yechimi:
42+4=46 g Em
100+15=115 g Em 115 g ---------15 g tuz 46 g -----------x=6 g tuz 6-4=2 g tuz 42 g eritmada 2/42×100=4,76% J: A
4. Xg 20% li H2SO 4 eritmasiga dastlab 100 gr suv .songra y gr 40% li H2 SO4 eritmasi qo'shildi.Natijada 1000 gr 20% li H2SO 4 eritmasi hosil bo'ldi.x gr 20%li H2SO4 eritmasiga 100 g suv qo'shilganda necha% li eritma hosil bo'ladi. A)5 B)8 C)12 D)16 Yechimi:
+Y g 40% H2SO4 1000×0,2=200 g kislota 1000-100=900 g eritma 0,2x+0,4y=200 X+Y=900 X=800 g. Y=100 g 800×0,2=160 g kislota 160
800+100 J: 17,78% 5. 400 g x %li KOH eritmasi bilan 600 g y %li LiOH eritmalari aralashtirildi.Agar hosil bo'lgan eritmadagi ishqorlarning massa ulushlari yig'indisi 0,136 ga teng bo'lsa,x va y larni(%) toping.(y%-x%=6). A)6;12
C)14;20 D)16;22 Yechimi:
4x+6y
y-x=6
x=10. y=16
J: B
6. 200 g x-% li NaCl eritmasiga uning x g x-% li eritmasi qo’shildi. Bunda 20% li osh tuzi (p = 1,1 g/ml) eritmasi hosil bo’ldi. Hosil bo’lgan eritmaning titrini (g/ml) hisoblang.
. x•x 2x+ ------- . 100 --------------------=0,2 200+x x=20
200•0,2=40 g NaCl
Em 200+20=220 g EV 220/1,1=200 ml 200 ml ------------44 g 1 ml----------x=0,22 g/ml J:0,22 g/ml 7. 100 gr X% eritmaga (y) gr tuz qowilganda (X+10)% li eritma xosil boldi.Hosil bo'lgan (X+10) % li eritmaga (2,5y) gr suv qo'shilganda (X-10) li eritma xosil bo'ldi. y ni aniqang Yechimi:
----------×100=x+10 100+y x+y
(100+y)+2,5y x=50. y=25 J: 25
8. 400 gr X% li KOH eritmasi bilan 600 gr Y% li LIOH eritmalari aralashtirildi. Agar hosil bo' lgan eritmadagi ishqorlarning massa ulush yig'indisi 0,18 ga teng bo' lsa X va Y larni(%) toping .(Y%-X%=6)
Yechimi:
400x+600y
Y-X=0,06
x=0,144. y=0,204
J: D
9. X mol ( suvda eruvchan ) Me2SO4 va 208 g BaCl2 tutgan eritmalari 500 g dan aralashtirilganda Y g 31,55% li eritma olindi. Me toping.
Me2SO4+BaSO4=2MeCl+BaSO4 BaCl2. BaSO4 208 g ----------------233 g 208 g ---------------x=233 g Em 500+500-233=767 g MeCl(m) 767×0,3155=242 g 2MeCl. 242/2=121 -35,5=85,5 g/ mol. Rb J: D
10. 800 gr 40% li Na2SO4 eritmasiga 200 gr suv qo'shilda va Y gr to'kib tashlandi va Y gr suv qo'shildi, va shu ketma ketlik 4 marta bajarilgan bo'lsa, 13,1072% eritma hosil bo'ldi. 3-marta to'kilayotganda necha gr tuz eritmadan chiqib ketadi ?
Yechimi: 800×0,4=320 g tuz 800+200=1000 g eritma 320×100/1000=32% eritma hosil bo'lgan Demak jarayon faqat 4 marta qaytarilgan desak 13,1071/32=0,4096 x^4=0,4096 √√0,4096 x=0,8 1-0,8=0,2×1000=200 g Y Demak: eritma 1000 g 32% I-bosqich.... Avval 200 g to'kildi 200×0,32=64 g tuz to'kildi . 320-64
. 1000-200 Endi SUV qo'shildi 256
800+200 II-bosqich Endi 200 g eritma to'kamiz 200×0,256=51,2 g tuz to'kildi Endi SUV qo'shamiz 204,8
800+200 III-bosqich Endi eritma to'kamiz 200×0,2048=40,96 g tuz to'kildi Endi suv qo'shamiz 163,84 ---------------×100=16,384% 800+200 IV-bosqich Endi eritma to'kamiz 200×0,16384=32,768 g tuz tokildi Endi SUV qo'shamiz 131,072
800+200 Jami chiqib ketgan tuzlarlarni hisoblaymiz I-64 g, II-51,2 g, III-40,96, IV-32,768 g Jami: 188,928 g III-40,96 g 11. X % li sulfat kislota eritmasiga uning massasidan 3 marta ko’p massadagi x-32,65 % li oleum qo’shilganda 84 % li eritma hosil bo’ldi. Oleum formulasini aniqlang? A) 4H2SO4∙0.1SO3 B) 5H2SO4∙1.25SO3 C) 6H2SO4∙1.25SO3 D) 4H2SO4∙1.35SO3 Yechimi: H2SO4 eritmasini 100 g x% Oleumni esa 300 g x-32,5% deb olamiz.... Hosil bo'lgan eritmadagi H2SO4 (m) aniqlaymiz 100+300=400 g eritma 400×0,84=336 g H2SO4 336 g H2SO4 uchun dastlabki moddalar necha (m) olinganini belgilab olamiz..... Dastlabki H2SO4 (m) H2SO4 100x/100=x g Endi oleum tarkibini belgilab olamiz... SO3+H2O=H2SO4 80 g 98 g 98/80=1,225 SO3 hosil bo'lgan H2SO4 aniqlaymiz (m) 300/100=3 deb olamiz (x-32,5)×3×1,225 Oleum tarkibidagi H2SO4( m) aniqlab olamiz... (100-(x-32,5))×3 Endi umumiy tenglama tuzamiz... 336=x+(100-(x-32,5))3+(x-32,5)3×1,225 x=34,6
H2SO4•xSO3=1+xH2SO4
98+80x. 98+98x --------------=---------------- 300. 336-34,6 x=0,0259
H2SO4•0,0259 SO3
4H2SO4•0,1SO3 J: A Yechimni tekshiramiz: 400-336=64 g suv qolgan dastlabki eritmadan 100-34,6-64=1,4 g suv sarflangan..... Dastlabki oleum tarkibidagi SO3 ni (m) aniqlaymiz... (x-32,5)×3 orqali (34,6-32,5)×3=6,3 g SO3 Endi sarflangan SUV massasini aniqlab taqqoslab ko'ramiz.. SO3+H2O=H2SO4 80 g SO3-------18 g SUV 6,3 g ------------x=1,4175 g SUV 1,4175-1,41=0,0075 g farq Yaxlitlashlar hisobiga chiqdi... 12. 100 g 20% li NaOH eritmasiga X g Na qo'shildi va Y g 40% li NaOH eritmasi hosil bo'lsa X va Y ni toping. A)19,15: 110,25 B)12: 117,15 C)10,9: 118,25 D)14,17: 114,1 Yechimi: NaOH(m) 100×0,2=20 g 2Na+2H2O=2NaOH+H2 46x. 80x. 2x 20+80x
(100+46x)-2x x=0,32
Na(m) 0,32×46=14,72 g
Em=100+14,72-0,64=114,08 g J: D
13. X% li eritma massasining 1/5 qismi bug’latildi. Bunda erituvchining massasi ¼ qisimga kamaydi. Eritmaning massa ulushi 5/4 marta ortsa, x ni aniqlang. Yechimi: (x/75+ 0,25x)= 1,25x/100 x= 20. Javob: D) 20 14.
Oleum 1. xH2SO4 ∙ ySO3 tarkibli oleumni netrallash uchun 110,5 ml (p=1,328) NaOH sariflandi. Hosil bolgan 40% li eritma tarkibida 0,55 M erigan modda bolsa oleum tarkibidagi SO3 ning massa ulushini (%) da niqlang. J: 49.5 % Yechimi: m(NaOH)=110,5*1,328=146,75 m(Na2SO4)=0,55*142=78,1/0,40=195,25 m(oleum)=195,25-146,75=48,5 x+y=0,55
x=0,25. y=0,3*80=24 24/48,5=0,49,5. yoki 49,5% 2. Massasi 3539,4 g suvda 15,6 g oleum eritilib 5 A tok bilan 57900 sekund davomida elektroliz qilingandan so’ng hosil bo’lgan eritmaning (ρ=1,05 g/ml) pH qiymati 1 ga teng bo’ldi. Oleum formulasini aniqlang. A) H2SO4·0,2SO3 C) H2SO4·0,5SO3 B) H2SO4·0,4SO3 D) H2SO4·0,8SO3 Yechimi:
m=---------=-------------------=27 g suv . F. 96500 Em=3539,4+15,6-27=3528 g EV 3528:1,05=3360 ml pH=1. [H^+] ant =0,1 mol/l 2H^+=H2SO4 2 mol--------1 mol 0,1 mol------x=0,05 mol/l 1000 ml-------0,05 mol 3360 ml---------x=0,168 mol H2SO4(m) 0,168×98=16,464 g H2SO4•xSO3=1+xH2SO4 98+80x. 98+98x -------------=-------------- 15,6. 16,464 x=0,4
J: H2SO4•0,4SO3
3. Massasi 70,8 g xH2SO4∙ySO3 tarkibli oleumga 180 g SO3 qo‘shilganda yH2SO4∙xSO3 tarkibli oleum hosil bo‘ldi. Boshlang‘ich oleumning 5,9 g massasini to‘la netrallash uchun KOH ning 43,75 g eritmasi sarflandi. Ishqor eritmasining molyal (mol/kg) konsentratsiyasini toping. A) 5,6 B) 3,4 C) 8,4 D) 16,4 Yechimi: xH2SO4•ySO4 yH2SO4•xSO3 SO3(n) 180/80=2,25 mol H2SO4(m) 98×2,25=220,5 g 98x. 70,8-80x --------------=----------------- 70,8-80x. 98x+220,5 x=0,15
SO3(m) 80×0,15=12 g
0,6:0,15
H2SO4•0,25SO4=1,25H2SO4 118 g -----------------122,5 g 5,9 g------------------x=6,125 g H2SO4(n) 6,125/98=0,0625 mol H2SO4+2KOH= 1 mol----------------2 mol 0,0625 mol------x=0,125 mol KOH(m) 0,125×56=7 g H2O(m) 43,75-7=36,75 g 36,75 g suv -----------0,125 mol 1000 g suv-----------x=3,4 mol J: 3,4 molyalli 4. Oleumga o'zining massasidan 1,5 marta kam suv qo'shilganda 65,4% li H2SO4 eritmasi hosil bo'ldi. Oleum tarkibidagi SO3 ning massa ulushini (%) toping? A)30 B)50 C)40 D)25 Yechimi:
100 g oleum 100/1,5=66,67 g SUV Em 100+66,67=166,67 g H2SO4 (m) 166,67×0,654=109 g H2SO4•xSO3=1+xH2SO4 98+80x. 98+98x --------------=-------------- . 100. 109 x=0,82
H2SO4•0,82SO3
0,82×80
98+(0,82•80) J:40
5. Tarkibi qanday bo'lgan oleumga 0,2 mol suv qo'shilganda sp2 va sp3 orbitallar soni 5:2 nisbatda bo'lgan 103,2 g oleum olindi oleum tarkibini toping.
Yechimi: H2SO4 sp3=4×3=12 ta (x), sp2 =2×3=6 ta (x) SO3 da sp3=0, sp2=4×3=12 ta (y) 98x+80y=103,2 6x+12y. 5 -------------=-------- . 12x. 2 x=0,4. y=0,8 SO3 + H2O = H2SO4 1 mol--1 mol ---1 mol 0,2=x-----0,2-------x=0,2 H2SO4 0,4-0,2=0,2 mol---1 SO3 0,8+0,2=1 mol---x=5 J: D
6. 27,6 g 29% li oleum 7,15 g kristall soda va 31,2 g 5% li natriy gidrosulfit aralashtirildi. Hosil bo'lgan eritmadagi erigan moddalarning massa ulushini toping. A)0,48 B)0,52 C)0,6 D)0,55 Yechimi: 27,6×0,29=8 g SO3 8/80=0,1 mol SO3 27,6-8=19,6/98=0,2 mol H2SO4 0,2H2SO4•0,1SO3=0,3H2SO4 Na2CO3•10H2O (n) 7,15/286=0,025 mol NaHSO3(n) 1,56/104=0,015 mol Na2CO3•H2O+H2SO4=Na2SO4+CO2+11H2O 1 mol-------------1 mol H2SO4 0,025 mol------x=0,025 mol 1 mol-----------1 mol Na2SO4 0,025 mol------x=0,025 mol 1 mol-------------44 g CO2 0,025 mol------x=1,1 g 2NaHSO3+H2SO4=Na2SO4+2SO2+2H2O 2 mol-------------1 mol H2SO4 0,015 mol------x=0,0075 mol 2 mol---------------1 mol Na2SO4 0,015 mol------x=0,0075 mol 2 mol-------------128 g SO2 0,015 mol------x=0,96 g H2SO4(m) 0,3-0,025-0,0075=0,2675×98=26,15 g Na2SO4(m) 0,025+0,0075=0,0325×142=4,615 g Em= (27,6+7,15+31,2)-(1,1+0,96)=63,89 g 26,215+4,615 -----------------------=0,4825 . 63,89 J: A
7. Noma’lum massadagi sulfat kislota eritmasini to’la neytrallash uchun 300 gr 80% li NaOH eritmasi sarflandi. Huddi shunday massali oleum eritmasini to’la neytrallash uchun esa 690 g 56% li KOH eritmasi sarflandi. Agar dastlabki oleum va sulfat kislota eritmalari teng massada aralashtirilganda H2SO4•0,29SO3 tarkibli oleum hosil bo’lishi ma’lum bo’lsa, dastlabki sulfat kislota eritmasining foiz konsentrasiyasini aniqlang. Yechish: 1) 300 g (80%)=240 g 240/40= 6 mol 2NaOH +H2SO4--> 2 mol ----- 1 mol 6 mol ----×=3 mol 2) 690 g (56%)=386,4 g 386,4/56=6,9mol 2KOH + OLEUM --> 2 mol ------ 1 mol 6,9 mol ----×=3,45 mol. 3) 3 + 3,45 = 6,45 mol H2SO4 * 0,29SO3 1,29 mol ------- 121,2 g 6,45 mol -----x= 606 g /2=303 g W=294/303=97% |
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