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Molecular Biology - Water

Grade 11IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Water molecules (H2OH_2O) are formed by polar covalent bonds between one oxygen atom and two hydrogen atoms, resulting in a dipole.

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The oxygen atom is more electronegative, attracting electrons more strongly and acquiring a partial negative charge (Ξ΄βˆ’\delta^-), while hydrogen atoms acquire a partial positive charge (Ξ΄+\delta^+).

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Hydrogen bonding is the intermolecular force that forms when the Ξ΄+\delta^+ hydrogen of one water molecule is attracted to the Ξ΄βˆ’\delta^- oxygen of another.

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Cohesion is the tendency of water molecules to stick to each other due to hydrogen bonding, which creates surface tension.

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Adhesion is the attraction of water molecules to other polar or charged surfaces, essential for capillary action in the xylem of plants.

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Thermal properties include a high specific heat capacity (4.186 Jβ‹…gβˆ’1β‹…βˆ˜Cβˆ’14.186 \, J \cdot g^{-1} \cdot ^\circ C^{-1}), meaning it takes significant energy to change water's temperature.

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High latent heat of vaporization (2257 Jβ‹…gβˆ’12257 \, J \cdot g^{-1}) allows water to act as an effective coolant through evaporation (e.g., sweating).

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Solvent properties: Water is the 'universal solvent' for polar and ionic substances (e.g., Na+Na^+, Clβˆ’Cl^-, glucose) because its dipoles surround and isolate the solute particles.

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Methane (CH4CH_4) vs. Water (H2OH_2O): Methane is non-polar and does not form hydrogen bonds, leading to a much lower boiling point (βˆ’161∘C-161^\circ C) compared to water (100∘C100^\circ C).

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Hydrophilic substances are chemically attracted to water (polar), while hydrophobic substances (non-polar like lipids) are repelled by it.

πŸ“Formulae

Q=mβ‹…cβ‹…Ξ”TQ = m \cdot c \cdot \Delta T

Q=mβ‹…LvQ = m \cdot L_v

Density (ρ)=mV\text{Density } (\rho) = \frac{m}{V}

H2Oβ‡ŒH++OHβˆ’H_2O \rightleftharpoons H^+ + OH^-

πŸ’‘Examples

Problem 1:

Compare the thermal properties of Water (H2OH_2O) and Methane (CH4CH_4) and explain the biological significance.

Solution:

Water has a boiling point of 100∘C100^\circ C and a specific heat capacity of 4.18 Jβ‹…gβˆ’1β‹…βˆ˜Cβˆ’14.18 \, J \cdot g^{-1} \cdot ^\circ C^{-1}. Methane has a boiling point of βˆ’161∘C-161^\circ C and a specific heat capacity of 2.2 Jβ‹…gβˆ’1β‹…βˆ˜Cβˆ’12.2 \, J \cdot g^{-1} \cdot ^\circ C^{-1}.

Explanation:

Because H2OH_2O is polar and forms hydrogen bonds, it requires much more energy to break these bonds during phase changes or temperature increases. This allows water to remain liquid in most habitats on Earth and act as a stable medium for metabolic reactions.

Problem 2:

Calculate the energy required to evaporate 50 g50 \, g of sweat from the skin, given the latent heat of vaporization of water is 2257 Jβ‹…gβˆ’12257 \, J \cdot g^{-1}.

Solution:

Q=mβ‹…Lv=50 gΓ—2257 Jβ‹…gβˆ’1=112,850 JQ = m \cdot L_v = 50 \, g \times 2257 \, J \cdot g^{-1} = 112,850 \, J

Explanation:

This large amount of energy is absorbed from the body's skin (heat of vaporization), which results in a significant cooling effect for the organism.

Problem 3:

Explain how Sodium Chloride (NaClNaCl) dissolves in water using LaTeX notation for the ions.

Solution:

Water molecules surround the ions; the Ξ΄βˆ’\delta^- oxygen atoms face the Na+Na^+ ions, and the Ξ΄+\delta^+ hydrogen atoms face the Clβˆ’Cl^- ions.

Explanation:

The formation of 'hydration shells' around Na+Na^+ and Clβˆ’Cl^- prevents the ions from re-associating, keeping the salt in a dissolved state.

Water - Revision Notes & Key Diagrams | IB Grade 11 Biology