Masalalar echishga doir namunalar




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Masalalar echishga doir namunalar

1-masala. 20 % li NaOH eritmasi tayyorlash uchun 2 mol’ NaOH ni qancha hajm suvda eritish kerak.



Berilgan:

C% = 20 %



n =2 mol



Echish. 1) 100 g eritmada 20 g NaOH va 80 g (d=1 g/ml bo`lgani sababli 80 ml) H2O bo`ladi.

2) 20 g NaOH da necha mol’ bor.

MNaOH= 40 g/mol’

nNaOH =20/40 =0,5 mol’

3) 2 mol NaOH qancha suvda eritilishi kerak.

80 g H2O --- 0,5 mol’ NaOH

X g H2O --- 2 mol’ NaOH

Demak, 20 % li eritma tayyorlash uchun 2 mol’ NaOH ni 320 ml H2O ga eritish kerak.

2-masala. 270 ml H2O 30 g NaCl eritildi. Hosil bo`lgan eritmani foiz(%) kontsentratsiyasi hisoblang.

Berilgan:

m = 30 g = 30*10-3kg = 3*10-2 kg



C% - ?



Echish. eritmani foiz(%) kontsentratsiyasi quyidagi formula bilan hisoblanadi:

bo`lgani sababli m =d*V = 103 *2,7*10-4 = 0,27 kg



3-masala. 10 % li 0,5 l HCl eritmasi tayyorlash uchun 30 % li eritmasidan necha ml olish kerak.



Berilgan:

V = 0,5 l = 500ml

C% = 10 %

C% = 30 %

V - ?


Echish. 1) 10 % li 500 ml HCl eritmasing massasi hisoblanadi:

10 % eritmani zichligi d =1,05 g/ml



2) Shu eritmadagi HCl ning massasi hisoblanadi:

100 - 10 g HCl

525 - X g HCl



3) 52,5 g HCl 30 % eritmaning qanchasi buladi?

30 % eritmaning zichligi d = 1,15 g/ml

100 - 30 g HCl

X - 52,5 g

4) 175 g HCl saqlagan 30 % eritmani hajmini hisoblanadi:



4-masala.10 % li HCl eritmasing molyar kontsentratsiyasi hisoblang (d =1,05 g/ml).



Berilgan:

C% = 10 %



d =1,05 g/ml= 1,05*10-3 kg/m3

Cm - ?



Echish. 1) 10 % li 1 l eritmani massasi hisoblanadi:

m=V*d=1000*1,05 =1050 g = 1,05 kg

2) 1050 g 10 % HCl eritmasidagi HCl ning massasi hisoblanadi:

100 --- 10 g HCl

1050 --- x g HCl

3) Modda miqdori «mol»larda hisoblanadi:



5-masala. 2 l 0,5 m H2SO4 eritmasi tayyorlash uchun, 10 % li H2SO4 dan necha ml olish kerak.



Berilgan:

V = 2 l = 2000 ml

Cm = 0,5 m

C% = 10 %

V - ?


Echish. 1) 2 l 0,5 m eritmada necha gramm H2SO4 borligi hisoblanadi:

1 l – 49 g H2SO4 – 0,5 m

2 l - x g H2SO4 – 0,5 m

X = 2 * 49 = 98 g = 0,098 kg

2) 10 % eritmani qanchasida 98 g H2SO4 bo`lishini topamiz:

100 g --- 10 g H2SO4

y g --- 98 g H2SO4

3) 10 % eritmani hajmini hisoblanadi:



6-masala. 5 l 2 n H3PO4 eritmasi tayyorlash uchun 30 % (d=1,18 g/ml) eritmadan qancha hajm olish kerak.



Berilgan:

V = 5 l h 5*10-3m3

Cn = 2 n

C% = 30 %

V - ?


Echish. 5 l 2 n H3PO4 eritmasida necha g-ekvivalent molyar massa bo`ladi.

;

1 l --- 32,66 g/mol’ H3PO4

5 l --- x g H3PO4

X = 5*2*32,66 = 326,6 g = 0,3266 kg

2) 326,6 g H3PO4 30 % eritmaning qancha miqdorida bo`ladi.

100 g --- 30 g H3PO4

X --- 326,6 g H3PO4

3) Kislota hajmi hisoblanadi:



7-masala. Kislotali muhitda ishlatiladigan 0,5 n 2 l KMnO4 eritmasi tayyorlash necha gramm KMnO4 kerak bo`ladi.



Berilgan:

Cn =0,5 n



V = 2 l = 2000 ml = 2*10-3m3

m - ?


Echish. 1) KMnO4 ning kislotali muhit uchun ekvivalent molyar massasi hisoblanadi.

MnO-4 + 8H+ + 5e → Mn2+ + 4 H2O



2) 0,5 n 2 l eritma tayyorlash uchun kerak bo`lgan KMnO4 massasi hisoblanadi.







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