Lipotso tsa Mehlala le Puisano ea Tharollo ea ho Hlōleha ha Libaka tse sa Feleng
Ho tepella maikutlong ha ntlha e batang ke ketsahalo ea bohlokoa ea ho thulana k'hemistri ea tharollo. E susumetsa lits'ebetso tse ngata tsa baeloji, lik'hemik'hale le tsa indasteri. Sengoloa sena se tšohla mehlala e 'maloa ea ho tepella maikutlong ha ntlha e batang' me se fana ka lipuisano tse qaqileng ho hlakisa mohopolo ona.
Ho Utloisisa ho Tepella Maikutlo ha Libaka tse sa Lokang
Ho tepella ha ntlha ea leqhoa ke ketsahalo eo ntlha ea leqhoa ea solvent e theoloang ho eona ha ho eketsoa solute ho eona. Phello ena ke ketsahalo ea ntoa, ho bolelang hore ho tepella ha ntlha ea leqhoa ho itšetlehile ka palo ea likaroloana tse qhibilihang ka har'a tharollo, eseng ka boitsebiso ba solute.
Tekanyo e akaretsang e sebelisetsoang ho bala ho tepella ha ntlha ea leqhoa ke:
\[
\Delta T_f = K_f \cdot m
\]
Di mana:
– \(\Delta T_f\) ke ho tepella ha ntlha ya serame.
– \(K_f\) ke tšobotsi e sa fetoheng ea ntlha ea ho hoama ea solvent.
– \(m\) ke molality ea tharollo (li-moles tsa solute/kg ea solvent).
Lipotso tsa Mohlala oa ho Tepella Maikutlo ha Libaka Tse Hatsetsang
Potso ea 1
Hoa tsebahala:
– Tharollo e etsoa ka ho qhala digrama tse 5 tsa NaCl (boima ba molar = 58.5 g/mol) ka digrama tse 200 tsa metsi.
– Boemo ba ho fokotseha ha ntlha ya ho hatsela (K_f) ba metsi ke 1.86 °C •kg/mol.
Potso:
Bala tekanyo ea ho hatsela ha tharollo ena.
Puisano:
1. Bala molality ea tharollo (m):
Taba ea pele, re bala palo ea li-moles tsa NaCl:
\[
\text{mol NaCl} = \frac{\text{boima ba NaCl (g)}}{\text{boima ba molar (g/mol)}} = \frac{5 \text{ g}}{58.5 \text{ g/mol}} = 0.0855 \text{ mol}
\]
Ebe re bala molality (li-moles tsa solute/kg ea solvent):
\[
\text{molality} = \frac{\text{mol NaCl}}{\text{boima ba solvent (kg)}} = \frac{0.0855 \text{ mol}}{0.2 \text{ kg}} = 0.4275 \text{ m}
\]
2. Bala ho fokotseha ha ntlha ya ho hatsela (\(\Delta T_f\)):
\[
\Delta T_f = K_f \cdot m = 1.86 \text{ °C•kg/mol} \cdot 0.4275 \text{ m} = 0.79545 \text{ °C}
\]
Kahoo, tharollo ena e tla ba le ho fokotseha ha ntlha ea leqhoa la 0.79545 °C.
Potso ea 2
Hoa tsebahala:
– Tharollo e na le digrama tse 10 tsa glucose (C₆H₁₂O₆, boima ba molar = 180 g/mol) e qhibilihisitsoeng ka digrama tse 250 tsa metsi.
– Boemo ba ho fokotseha ha ntlha ea ho hatsela bakeng sa metsi ke 1.86 °C •kg/mol.
Potso:
Tharollo ena ea glucose e baka ho fokotseha ha ntlha ea ho hoamisa ke eng?
Puisano:
1. Bala molality ea tharollo (m):
Taba ea pele, re bala palo ea li-mole tsa glucose:
\[
\text{mol glucose} = \frac{\text{boima ba glucose (g)}}{\text{boima ba molar (g/mol)}} = \frac{10 \text{ g}}{180 \text{ g/mol}} = 0.0556 \text{ mol}
\]
Ebe re bala molality (li-moles tsa solute/kg ea solvent):
\[
\text{molality} = \frac{\text{mol glucose}}{\text{boima ba solvent (kg)}} = \frac{0.0556 \text{ mol}}{0.25 \text{ kg}} = 0.2224 \text{ m}
\]
2. Bala ho fokotseha ha ntlha ya ho hatsela (\(\Delta T_f\)):
\[
\Delta T_f = K_f \cdot m = 1.86 \text{ °C•kg/mol} \cdot 0.2224 \text{ m} = 0.413664 \text{ °C}
\]
Kahoo, tharollo ea tsoekere e tla ba le phokotso ea ntlha ea ho hoama ea 0.413664 °C.
Potso ea 3
Hoa tsebahala:
– Tharollo e etsoa ka ho qhala digrama tse 20 tsa urea (NH₂CONH₂, boima ba molar = 60 g/mol) ka digrama tse 500 tsa metsi.
– Boemo ba ho fokotseha ha ntlha ya ho hatsela (K_f) ba metsi ke 1.86 °C •kg/mol.
Potso:
Bala tekanyo ea ho hatsela ha tharollo ena.
Puisano:
1. Bala molality ea tharollo (m):
Taba ea pele, re bala palo ea li-mole tsa urea:
\[
\text{mole urea} = \frac{\text{boima ba urea (g)}}{\text{boima ba molar (g/mol)}} = \frac{20 \text{ g}}{60 \text{ g/mol}} = 0.3333 \text{ mol}
\]
Ebe re bala molality (li-moles tsa solute/kg ea solvent):
\[
\text{molality} = \frac{\text{mol urea}}{\text{boima ba solvent (kg)}} = \frac{0.3333 \text{ mol}}{0.5 \text{ kg}} = 0.6666 \text{ m}
\]
2. Bala ho fokotseha ha ntlha ya ho hatsela (\(\Delta T_f\)):
\[
\Delta T_f = K_f \cdot m = 1.86 \text{ °C•kg/mol} \cdot 0.6666 \text{ m} = 1.240476 \text{ °C}
\]
Kahoo, tharollo e tla ba le ho theoha ha ntlha ea leqhoa ea 1.240476 °C.
Potso ea 4
Hoa tsebahala:
– Tharollo e na le digrama tse 15 tsa asetiki (CH₃COOH, boima ba molar = 60 g/mol) e qhibilihisitsoeng ka digrama tse 300 tsa metsi.
– Boemo ba ho fokotseha ha ntlha ya ho hatsela (K_f) ba metsi ke 1.86 °C •kg/mol.
Potso:
Tharollo ena ea acetic acid e baka ho hatsela ha ntlha ea leqhoa joang?
Puisano:
1. Bala molality ea tharollo (m):
Taba ea pele, re bala palo ea li-moles tsa acetic acid:
\[
\text{mol of acetic acid} = \frac{\text{boima ba acetic acid (g)}}{\text{molar weight (g/mol)}} = \frac{15 \text{ g}}{60 \text{ g/mol}} = 0.25 \text{ mol}
\]
Ebe re bala molality (li-moles tsa solute/kg ea solvent):
\[
\text{molality} = \frac{\text{mol of acetic acid}}{\text{mass of solvent (kg)}} = \frac{0.25 \text{ mol}}{0.3 \text{ kg}} = 0.8333 \text{ m}
\]
2. Bala ho fokotseha ha ntlha ya ho hatsela (\(\Delta T_f\)):
\[
\Delta T_f = K_f \cdot m = 1.86 \text{ °C•kg/mol} \cdot 0.8333 \text{ m} = 1.55 \text{ °C}
\]
Kahoo, tharollo e tla ba le ho theoha ha ntlha ea leqhoa ea 1.55 °C.
Qetello
Ho tepella maikutlong ha ntlha e batang ke ketsahalo ea bohlokoa ea ho kopana le batho ba bang lits'ebetsong tse fapaneng tsa saense le tsa indasteri. Ka ho utloisisa mokhoa oa ho bala ho tepella maikutlong ha ntlha e batang, re ka bolela esale pele hore na litharollo tse fapaneng li tla ama ntlha e batang joang ea tharollo. Sengoloa sena se buile ka mehlala e 'maloa ea mathata le mokhoa oa ho bala ho tepella maikutlong ha ntlha e batang bakeng sa litharollo, ho u fa motheo o tiileng oa ho utloisisa mohopolo oa ho tepella maikutlong ha ntlha e batang.
