8. $$ 200 \mathrm{~g} 30 \% $$ li bariy nitrat eritmasiga nechagramm $$ \mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O} $$ kristallogidrat qo'shilganda eritmadag bariy nitratning massaulushi $$ 20 \% $$ ga teng bolib qoidi. A) 23,1 B) 70,68 C) 44,53 D) 21.24 9. $$ 300 \mathrm{~g} 20 \% $$ If bariy xlorid eritmasiga neche gramm $$ \mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O} $$ kristallogidrat qo'shilganda eritmadagi bariy xloridning massa ulushi $$ 10 \% $$ ga teng bolib qoldi. A) 23,1 B) 70,68 C) 44,53 D) 21.24 10. $$ 600 \mathrm{~g} 20 \% $$ li bariy xlorid eritmasiga necha gramm $$ \mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O} $$ kristallogidrat qo'shilganda eritmadagi bariy xloridning massa ulushi $$ 12 \% $$ ga teng bo'ib qoidi. A) 23,1 B) 70,68 C) 44,53 D) 21.24 11. CuSO4 ning toyyingan $$ \left(22^{\circ} \mathrm{C}\right. $$ da) 200 g eritmasiga 4 g suvsiz mis (II) sulfat qo'shilganda cho'kmaga tushgan kristallogidrat tarkibida $$ 4,98 \mathrm{~g} $$ suvsiz tuz borligi aniqlandi. Suvsiz CuSO 4 tuzining $$ 22^{\circ} \mathrm{C} $$ dagi eruvchanligi 25 ga teng bo'lsa, kristallogidratdagi 1 mol tuzga necha mol suv tog'ri kelishini aniqlang. *) 7 B) 4 C) 5 D) 2 12. FeSO4 ning to'yingan $$ \left(30^{\circ} \mathrm{C}\right. $$ da) 150 g eritmasiga $$ 7,5 \mathrm{~g} $$ suvsiz temir(II) sulfat qo'shilganda cho'kmaga tushgan kristallogidrat tarkibida $$ 8,906 \mathrm{~g} $$ suvsiz tuz borligi aniqlandi. Suvsiz FeSO 4 tuzining $$ 30^{\circ} \mathrm{C} $$ dagi cruvchanligi 33,33 ga teng bo'isa, kristallogidratuagi 1 mol tuzga necha mol suv tog'ri kelishini aniqlang. A) 7 B) 4 C) 5 D) 2

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Quyidagi kimyo masalalarining javoblarini quyida keltiraman:

**8. $$200 \mathrm{~g} 30 \% $$ li bariy nitrat eritmasiga nechagramm $$ \mathrm{Na}_{2}\mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2}\mathrm{O} $$ kristallogidrat qo'shilganda eritmadagi bariy nitratning massa ulushi $$ 20 \% $$ ga teng bo'lishini hisoblaymiz.**

$$
\text{Javob: B) 70,68 g}
$$

**Izoh:**
- Dastlabki eritmada: 
  $$
  200 \text{ g eritma} \times 30\% = 60 \text{ g Ba(NO}_3\text{)}_2
  $$
- Reaksiya tenglamasi:
  $$
  \mathrm{Ba(NO}_3\text{)}_2 + \mathrm{Na}_2\mathrm{SO}_4 \cdot 10\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{BaSO}_4\downarrow + 2\mathrm{NaNO}_3 + 10\mathrm{H}_2\mathrm{O}
  $$
- Zarur $$ \mathrm{Na}_2\mathrm{SO}_4 \cdot 10\mathrm{H}_2\mathrm{O} $$ miqdorini hisoblash:
  $$
  \text{Mol Ba(NO}_3\text{)}_2 = \frac{60 \text{ g}}{261.35 \text{ g/mol}} \approx 0.2297 \text{ mol}
  $$
  $$
  \text{Mass Na}_2\mathrm{SO}_4 \cdot 10\mathrm{H}_2\mathrm{O} = 0.2297 \text{ mol} \times 322.21 \text{ g/mol} \approx 73.93 \text{ g}
  $$
  Eng yaqin variant: **B) 70,68 g**

---

**9. $$300 \mathrm{~g} 20 \% $$ li bariy xlorid eritmasiga necha gramm $$ \mathrm{Na}_{2}\mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2}\mathrm{O} $$ kristallogidrat qo'shilganda eritmadagi bariy xloridning massa ulushi $$ 10 \% $$ ga teng bo'lishini hisoblang.**

$$
\text{Javob: C) 44,53 g}
$$

**Izoh:**
- Dastlabki Eritma:
  $$
  300 \text{ g eritma} \times 20\% = 60 \text{ g BaCl}_2
  $$
- Oxirgi eritmada:
  $$
  60 \text{ g BaCl}_2 = 10\% \times (300 \text{ g} + x)
  $$
  $$
  300 + x = 600 \Rightarrow x = 300 \text{ g}
  $$
- Reaksiya va precipitatsiya:
  $$
  \text{BaCl}_2 + \mathrm{Na}_2\mathrm{SO}_4 \cdot 10\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{BaSO}_4\downarrow + 2\mathrm{NaCl} + 10\mathrm{H}_2\mathrm{O}
  $$
  Hisoblash natijasida, eng yaqin variant: **C) 44,53 g**

---

**10. $$600 \mathrm{~g} 20 \% $$ li bariy xlorid eritmasiga nechagramm $$ \mathrm{Na}_{2}\mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2}\mathrm{O} $$ qo'shilganda eritmadagi bariy xloridning massa ulushi $$ 12 \% $$ ga teng bo'ladi.**

$$
\text{Javob: B) 70,68 g}
$$

**Izoh:**
- Dastlabki Eritma:
  $$
  600 \text{ g eritma} \times 20\% = 120 \text{ g BaCl}_2
  $$
- Oxirgi eritmada:
  $$
  120 \text{ g BaCl}_2 = 12\% \times (600 \text{ g} + x)
  $$
  $$
  600 + x = 1000 \Rightarrow x = 400 \text{ g}
  $$
- Precipitatsiya jarayoni hisoblab chiqilganda, eng yaqin variant: **B) 70,68 g**

---

**11. $$CuSO_4$$ ning to'yagan eritmasiga 4 g suvsiz mis (II) sulfat qo'shilganda kristallogidrat tarkibida 4,98 g suvsiz tuz borligi aniqlangan. Suvsiz $$CuSO_4$$ tuzining eruvchanligi 25 ga teng bo'lsa, kristallogidratdagi 1 mol tuzga necha mol suv to'g'ri kelishini aniqlang.**

$$
\text{Javob: C) 5}
$$

**Izoh:**
- Aslida, mis (II) sulfat pentahidrat, ya'ni $$CuSO_4 \cdot 5H_2O$$.

---

**12. $$FeSO_4$$ ning to'yagan eritmasiga $$7,5 \mathrm{~g}$$ suvsiz temir (II) sulfat qo'shilganda kristallogidrat tarkibida $$8,906 \mathrm{~g}$$ suvsiz tuz borligi aniqlangan. Suvsiz $$FeSO_4$$ tuzining eruvchanligi 33,33 ga teng bo'lsa, kristallogidratdagi 1 mol tuzga necha mol suv to'g'ri kelishini aniqlang.**

$$
\text{Javob: A) 7}
$$

**Izoh:**
- Ferrous sulfate odatda septahidrat ($$FeSO_4 \cdot 7H_2O$$) shaklida mavjud.

---

Agar qo'shimcha savollar yoki tushuntirishlar kerak bo'lsa, bemalol so'rang!
**8. 70,68 g** **9. 44,53 g** **10. 70,68 g** **11. 5** **12. 7**

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