Yechimlarning muzlash nuqtasi tushkunligi bo'yicha namunaviy savollar va muhokama
Muzlash nuqtasining pasayishi eritma kimyosida muhim kolligativ hodisadir. U ko'plab biologik, kimyoviy va sanoat jarayonlariga ta'sir qiladi. Ushbu maqolada muzlash nuqtasining pasayishining bir nechta misollari muhokama qilinadi va ushbu kontseptsiyani aniqlashtirish uchun batafsil muhokamalar keltirilgan.
Muzlash nuqtasi depressiyasini tushunish
Muzlash nuqtasining pasayishi - bu erituvchiga erigan modda qo'shilganda uning muzlash nuqtasi pasayishi hodisasi. Bu effekt kolligativ hodisa bo'lib, muzlash nuqtasining pasayishi eritmadagi erigan zarrachalar soniga bog'liq, erigan moddaning o'ziga xosligiga emas.
Muzlash nuqtasining pasayishini hisoblash uchun ishlatiladigan umumiy tenglama:
\[
\Delta T_f = K_f \cdot m
\]
Qayerda:
– \(\Delta T_f\) muzlash nuqtasi tushkunligidir.
– \(K_f\) erituvchining xarakterli muzlash nuqtasining tushkunlik konstantasidir.
– \(m\) eritmaning molalligi (erigan moddaning mollari/kg erituvchi).
Muzlash nuqtasidagi depressiyaga oid misol savollari
1-savol
Bu aniq:
– Eritma 5 gramm NaCl (molyar og'irlik = 58.5 g/mol) ni 200 gramm suvda eritish orqali tayyorlanadi.
– Suvning muzlash nuqtasining tushkunlik konstantasi (K_f) 1.86 °C•kg/mol ga teng.
Savol:
Ushbu eritmaning muzlash nuqtasining pasayishini hisoblang.
Munozara:
1. Eritmaning molalligini (m) hisoblang:
Birinchidan, biz NaCl mollarining sonini hisoblaymiz:
\[
\text{mol NaCl} = \frac{\text{NaCl massasi (g)}}{\text{molyar og'irlik (g/mol)}} = \frac{5 \text{ g}}{58.5 \text{ g/mol}} = 0.0855 \text{ mol}
\]
Keyin, biz molallikni (erigan moddaning mollari/kg erituvchi) hisoblaymiz:
\[
\text{molallik} = \frac{\text{mol NaCl}}{\text{erituvchining massasi (kg)}} = \frac{0.0855 \text{mol}}{0.2 \text{kg}} = 0.4275 \text{m}
\]
2. Muzlash nuqtasi tushkunligini hisoblang (\(\Delta T_f\)):
\[
\Delta T_f = K_f \cdot m = 1.86 \text{ °C•kg/mol} \cdot 0.4275 \text{ m} = 0.79545 \text{ °C}
\]
Shunday qilib, bu eritma 0.79545 °C muzlash nuqtasining pasayishini boshdan kechiradi.
2-savol
Bu aniq:
– Eritmada 250 gramm suvda erigan 10 gramm glyukoza (C₆H₁₂O₆, molyar og'irligi = 180 g/mol) mavjud.
– Suvning muzlash nuqtasining pasayishi konstantasi 1.86 °C•kg/mol ga teng.
Savol:
Bu glyukoza eritmasining muzlash nuqtasi qanday pasayadi?
Munozara:
1. Eritmaning molalligini (m) hisoblang:
Birinchidan, biz glyukoza mollari sonini hisoblaymiz:
\[
\text{mol glyukoza} = \frac{\text{glyukoza massasi (g)}}{\text{molyar og'irlik (g/mol)}} = \frac{10 \text{ g}}{180 \text{ g/mol}} = 0.0556 \text{ mol}
\]
Keyin, biz molallikni (erigan moddaning mollari/kg erituvchi) hisoblaymiz:
\[
\text{molallik} = \frac{\text{mol glyukoza}}{\text{erituvchining massasi (kg)}} = \frac{0.0556 \text{mol}}{0.25 \text{kg}} = 0.2224 \text{m}
\]
2. Muzlash nuqtasi tushkunligini hisoblang (\(\Delta T_f\)):
\[
\Delta T_f = K_f \cdot m = 1.86 \text{ °C•kg/mol} \cdot 0.2224 \text{ m} = 0.413664 \text{ °C}
\]
Shunday qilib, glyukoza eritmasining muzlash nuqtasi 0.413664 °C ga pasayadi.
3-savol
Bu aniq:
– Eritma 20 gramm karbamid (NH₂CONH₂, molyar og'irligi = 60 g/mol) 500 gramm suvda eritib tayyorlanadi.
– Suvning muzlash nuqtasining tushkunlik konstantasi (K_f) 1.86 °C•kg/mol ga teng.
Savol:
Ushbu eritmaning muzlash nuqtasining pasayishini hisoblang.
Munozara:
1. Eritmaning molalligini (m) hisoblang:
Birinchidan, biz karbamid mollari sonini hisoblaymiz:
\[
\text{mole karbamid} = \frac{\text{karbamid massasi (g)}}{\text{molyar og'irlik (g/mol)}} = \frac{20 \text{ g}}{60 \text{ g/mol}} = 0.3333 \text{ mol}
\]
Keyin, biz molallikni (erigan moddaning mollari/kg erituvchi) hisoblaymiz:
\[
\text{molallik} = \frac{\text{mol karbamid}}{\text{erituvchining massasi (kg)}} = \frac{0.3333 \text{mol}}{0.5 \text{kg}} = 0.6666 \text{m}
\]
2. Muzlash nuqtasi tushkunligini hisoblang (\(\Delta T_f\)):
\[
\Delta T_f = K_f \cdot m = 1.86 \text{ °C•kg/mol} \cdot 0.6666 \text{ m} = 1.240476 \text{ °C}
\]
Shunday qilib, eritma 1.240476 °C muzlash nuqtasining pasayishini boshdan kechiradi.
4-savol
Bu aniq:
– Eritmada 300 gramm suvda erigan 15 gramm sirka kislotasi (CH3₃COOH, molyar og'irligi = 60 g/mol) mavjud.
– Suvning muzlash nuqtasining tushkunlik konstantasi (K_f) 1.86 °C•kg/mol ga teng.
Savol:
Bu sirka kislotasi eritmasining muzlash nuqtasi qanday pasayadi?
Munozara:
1. Eritmaning molalligini (m) hisoblang:
Birinchidan, biz sirka kislotasining mol sonini hisoblaymiz:
\[
\text{sirka kislotasining mol miqdori} = \frac{\text{sirka kislotasining massasi (g)}}{\text{molyar og'irlik (g/mol)}} = \frac{15 \text{ g}}{60 \text{ g/mol}} = 0.25 \text{ mol}
\]
Keyin, biz molallikni (erigan moddaning mollari/kg erituvchi) hisoblaymiz:
\[
\text{molallik} = \frac{\text{sirka kislotasining mol miqdori}}{\text{erituvchining massasi (kg)}} = \frac{0.25 \text{ mol}}{0.3 \text{ kg}} = 0.8333 \text{ m}
\]
2. Muzlash nuqtasi tushkunligini hisoblang (\(\Delta T_f\)):
\[
\Delta T_f = K_f \cdot m = 1.86 \text{ °C•kg/mol} \cdot 0.8333 \text{ m} = 1.55 \text{ °C}
\]
Shunday qilib, eritma 1.55 °C muzlash nuqtasining pasayishini boshdan kechiradi.
Xulosa
Muzlash nuqtasining pasayishi turli ilmiy va sanoat qo'llanmalarida muhim kolligativ hodisadir. Muzlash nuqtasining pasayishini qanday hisoblashni tushunish orqali biz turli xil erigan moddalar eritmaning muzlash nuqtasiga qanday ta'sir qilishini oldindan aytib berishimiz mumkin. Ushbu maqolada bir nechta misol masalalar va eritmalar uchun muzlash nuqtasining pasayishini qanday hisoblash muhokama qilingan bo'lib, sizga muzlash nuqtasining pasayishi tushunchasini tushunishda mustahkam poydevor beradi.
