Fizika va kimyo

5-masala: NaOH suvdagi eritmasining molyal va foiz konsentratsiyalari nisbati 1:3,2 bo’lsa, eritmaning molyal konsentratsiyasini hisoblang.
nisbatni
deb belgilaymiz va formulaga qo’yamiz.


Javob: 6,25 Cm

6-masala: Bariy xlorid eritmasiga tegishli miqdorda Na₂SO4_*8H₂O kristallogidratidan qo’shilganda 93,3 g cho’kma tushdi va 2 molyalli eritma hosil bo’ldi. Dastlabki bariy xlorid eritmasining foiz konsentratsiyasini hisoblang.
Yechish: Cho’kma sifatida BaSO4 tushishini bilib,
n_(cho’kma)
m_(BaCl2)=0,4*208=83,2
m_(krist)=0,4*286=114,4 g
m_(NaCl)=0,4*2*58,5=46,8g

m_(eritma)=46,8/0,1047=447
m_(dast.eritm)=447+93,2=540,2-114,4=425,8 g

Javob: 19,54 %
7. Ikki valentli metall bo‘lagi xona haroratida xlorid kislotaning 54,6 g 10 molyalli eritmasida eritilganda H₂ gazi ajraldi va metall xloridining 32,2% li eritmasi hosil bo‘ldi. Hosil bo‘lgan metall xlorid molyalligini aniqlang.
Yechimi:

M_(HCl) = 10×36,5=365 g

M_(eritm) 365+1000=1365 g

1365 g --------------365 g HCl

54,6 g ---------------x=14,6 g

73 g 1 mol 1 mol 2g

2HCl + Me = MeCl2+ H2
14,6 0,2 0,2 0,4

. 0,2(x+71)

0,322=--------------------------
. (0,2x+54,6)-0,4

x=24 g/mol Mg

m_(MgCl2) =95*0,2=19 g

n_(MgCl2) 19/95=0,2 mol
m _(Mg)= 24*0,2=4,8 g
m _(E) =(4,8+54,6)-0,4=59 g
m _(H2O) =59-19=40 g

40 g suvda ---------0,2 mol

1000 g --------------x=5 mol