Фарғона туманидаги

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
2H2O=2H2+O2

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
NaOH(%)=------------------------×100=

. 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:
100 g suvda 80 g erigan

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:
X g 20%+100 g SUV

+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
---------------×100=17,78%

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
B)10;16

C)14;20
D)16;22

Yechimi:

4x+6y
---------------=0,136
400+600

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.
Yechimi:

. x•x
2x+ -------
. 100
--------------------=0,2
200+x

x=20

200•0,2=40 g NaCl
20•0,2=4 g NaCl
40+4=44 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:
x+y

----------×100=x+10
100+y

x+y
---------------------×100=x-10

(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)
A)14;20 B)10;16 C)16;22 D)14,4;20,4

Yechimi:

400x+600y
--------------------=0,18
400+600

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.
A)Li B)Na C)Cd D)Rb

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
%=-----------------×100=40%

. 1000-200

Endi SUV qo'shildi

256
---------------×100=25,6%

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
---------------×100=20,48%

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
---------------×100=13,1072%

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
1:0,0259. |4
4:0,104

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
----------------------=0,4

(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
98x+80y=48,5

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:
. J M t. 5×9×57900

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
H2SO4(m) 70,8-12=58,8 g
H2SO4(n) 58,8/98=0,6 mol

0,6:0,15
1:0,25

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:
Dastlabki oleumni 100 g deb olamiz

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
---------------------×100=40%

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.
A)2SO3•H2SO4 B)4SO3•H2SO4 C)0,25SO3•H2SO4 D)5SO3•H2SO4

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%

8.