Javob: 10"C da tuzning eruvchanlik koeffitsienti 9 gr ga teng.
2-misol. 46 g X % li eritmaga 2,8 g tuz qo’shilganda to’yingan eritma hosil bo’ldi. Shu tuzning eruvchanligi 22 ga teng bo’lsa, X ning qiymatini aniqlang.
Yechish: S=22
Eruvchanlik koeffitsiyentiga ko’ra 100 gr suvda 22 gr tuz eriganda 122 gr to’yingan eritma hosil qilgan. Tarkibida C%= mtuz/mer=22/122=0,1803 *100%=l 8,03% tuz
bor.To’yingan eritmaning o’zgarmas haroratda C % ham o’zgarmas bo’ladi degan qoidadan quyidagi tenglamani tuzamiz:
C%
m (tuz)
m (eritma)
0,1803 =
m (tuz)
46+ 2.8
To’yingan eritmadagi tuzning massasi m(tuz)=8,8 gr ga teng ekan, demak boshlang’ish eritmada mbosh(tuz)= 8,8-2,8=6gr tuz bo’lgan.
Uning C% = * 100% = 100% = 13,04% ga teng ekan.
m(eritma) 46
Javob: 13,04%
3-misol. Na2SO4 to’yingan eritmasiga Na2SO4 qo' shib qizdirildi. Bunda 500 g 20 % li eritma olindi. Dastlabki sharoitga keltirilganda 124,6 g Na2SO4.10H2O cho' kmaga tushdi. Dastlabki eritma (%) toping.
Yechish:
Masala shartiga ko’ra umumiy tuzning massasi 500 gr ni 20% ni tashkil qilyabdi, ya’ni
m(Na2SO4)=
G%* m ( eritma]
100%
500*20
100%
lOOgr.
Umumiy m(suv)=500-100=400gr. Eritmani qizdirib sovitish natijasida
Na2SO4*10H2O Na2SO4+ 10H2O
322 gr 142 gr 180gr
124.6gr x y
x= 55gr tuz y=69,6 gr suv kristallogidratga aylanib cho’kmaga o’tgan. Natijada eritmada m(suv)=400-69,6=330.4 gr qolgan,
m(tuz)= 100-55=45 gr qolgan.
45
To’yingan eritmaning o’zgarmas haroratda C % ham o’zgarmas bo’ladi.Ya’ni harorat ko’tarilib so’ng boshlang’ich haroratga keltirilgandan keyin eritmaning C% ham o’zearmavdi:
C% = m〔tUZ)、* 100% =* 100% = 12% m (eritma) 45 + 330,4
Javob:Demak boshlang’ich eritmada to’yingan eritmaning 12 %ini tuz tashkil qilar ekan.
Foyfalanilgan adabiyotlar ะ
O’zb eki st on Pespublikasi Prezidenti Sh.Mirziyayevning Oliy majlisga murojatnomasi;
I.R.ASKAROV,K.G’OPIROYN.X.TO’XTÂBOYEV KIMYO-8 darslik.TOSHKENT ‘ YANGIYUL POLIGRAPH SERVİCE” 2019
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