Problem 80 - Chemistry and Chemical Reactivity

Question

A tree is \(10.0 \mathrm{m}\) tall. (a) What must be the total molarity of the solutes if sap rises to the top of the tree by osmotic pressure at \(20^{\circ} \mathrm{C} ?\) Assume the groundwater outside the tree is pure water and that the density of the sap is \(1.0 \mathrm{g} / \mathrm{mL} .\left(1 \mathrm{mm} \mathrm{Hg}=13.6 \mathrm{mm} \mathrm{H}_{2} \mathrm{O} .\right)\) (b) If the only solute in the sap is sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11},\) what is its percent by mass?

Step-by-Step Solution

Answer

(a) Molarity is 0.040 mol/L; (b) Sucrose percent by mass is 1.369%.

Step 1: Convert tree height to water pressure

The tree's height is 10.0 m, which corresponds to a pressure due to the height of a column of water. Use the conversion given: \(1 \text{ mm Hg} = 13.6 \text{ mm } \text{H}_2\text{O}\). First, find the water pressure in terms of mm water:\[ 10.0 \text{ m} = 10,000 \text{ mm} \].

Step 2: Convert water pressure to pressure in mm Hg

Using the conversion factor, convert the tree height in mm water into mm Hg to find the osmotic pressure required: \[ \frac{10,000 \text{ mm}}{13.6} = 735.3 \text{ mm Hg}\].

Step 3: Use osmotic pressure formula to find molarity

The formula for osmotic pressure \( \Pi \) is given by \( \Pi = iMRT\), where \(i\) is the van't Hoff factor (1 for non-electrolyte), \(M\) is molarity, \(R\) is the ideal gas constant (0.0821 L atm mol⁻¹ K⁻¹), and \(T\) is temperature in Kelvin (20°C = 293K). First, convert the pressure into atm: \[ \frac{735.3}{760} = 0.968 atm \]. Use the formula to solve for \(M\): \[ M = \frac{\Pi}{RT} = \frac{0.968}{0.0821 \times 293} = 0.040 \text{ mol/L} \].

Step 4: Calculate percent by mass of sucrose

The molarity of 0.040 mol/L corresponds to sucrose with a molar mass of 342.3 g/mol. Calculate mass and then convert it to percent by mass. If density is 1.0 g/mL (or 1000 g/L), compute the mass of sucrose: \[ 0.040 \times 342.3 = 13.692 \text{ g} \].Percent by mass: \[ \left(\frac{13.692}{1000} \right) \times 100 \approx 1.369\% \].

Key Concepts

Molarity

Molarity is a measure of the concentration of a solute in a solution. It is expressed as the number of moles of solute per liter of solution. In the given exercise, we calculated the molarity needed to raise the sap to the top of a 10-meter-tall tree using osmotic pressure. Molarity (\(M\)) is calculated by dividing the osmotic pressure (\(\Pi\)) by the product of the gas constant (\(R\)), temperature (\(T\)), and the van't Hoff factor (\(i\)):

• Osmotic Pressure (\(\Pi\)): Measures how much pressure is needed to prevent osmosis. In this problem, it is converted from mm Hg to atm.
• Van't Hoff factor (\(i\)): For non-electrolytes like sucrose, \(i\) is 1.
• Temperature (\(T\)): Always in Kelvin for calculations. For 20°C, this is 293 K.
• Gas Constant (\(R\)): 0.0821 L atm mol⁻¹ K⁻¹.

Observing the formula \( M = \frac{\Pi}{RT} \) helps us determine how concentrated a solution must be to exert a specific osmotic pressure.

Percent by Mass

Percent by mass is a way of expressing the concentration of a component within a mixture or solution. It is calculated by taking the mass of the solute, dividing it by the total mass of the solution, and multiplying by 100 to get a percentage. For sucrose in the sap of the tree, percent by mass provides insight into how much of the sap's mass is due to sucrose.

• The mass of sucrose: Calculated using its molarity (moles per liter) and molar mass, which is 342.3 g/mol.
• Total mass of the solution: This refers to the mass of 1 liter of sap, which, given its density, is 1000 g/L.

After finding the mass of the sucrose in 1 liter of sap, divide this by the mass of the entire liter of sap and multiply by 100 to get the percent by mass. This simplification helps understand the composition of the sap in terms of the solute.

Sucrose

Sucrose is a common disaccharide sugar comprised of glucose and fructose. It has the molecular formula \(\text{C}_{12}\text{H}_{22}\text{O}_{11}\) and a molar mass of 342.3 g/mol. In the process of analyzing sap, sucrose acts as the solute that creates osmotic pressure allowing the sap to investigate osmosis.
Sucrose impacts sap characteristics such as:

• Osmotic pressure: As sucrose is a non-electrolyte, the van't Hoff factor is 1, simplifying the calculations of osmotic pressure.
• Dilution: The solubility of sucrose affects how it behaves in water, and understanding its concentration helps in hydration management in plants.

Thus, considering sucrose's molecular properties and interaction in a solution is crucial for understanding the osmotic behaviors in plants.

Van't Hoff Factor

The van’t Hoff factor (\(i\)) indicates the number of particles into which a solute disassociates in solution. For non-electrolytes like sucrose, \(i\) equals 1 because they do not dissociate into more particles in solution. This makes it simpler to calculate the osmotic pressure since the factor does not alter substantially.
When solving for osmotic pressure, the factor directly influences:

• Osmotic pressure (\(\Pi = iMRT\)): Higher \(i\) values increase pressure for a given concentration, but for non-dissociating sucrose, it stays simple.
• Colligative properties: \(i\) impacts boiling point elevation, freezing point depression, and osmotic pressure.

Although its role might seem understated in this scenario, the van’t Hoff factor can be essential when dealing with electrolytes, highlighting how non-electrolytes make for straightforward calculations in plant osmosis.

Osmosis

Osmosis is the movement of solvent molecules, typically water, through a semi-permeable membrane from a region of lower solute concentration to one of higher solute concentration. It is crucial for maintaining fluid equilibrium in biological systems.
In plants, osmosis assists in the transport of nutrients and water from the roots through the stem to the leaves. The osmosis process in the exercise occurs in the context of sap rising through the tree.

• Driving force: The difference in solute concentration between the groundwater and the sap generates osmotic pressure, causing water to move into the tree.
• Pressure generation: Osmotic pressure results from the solution and solute properties, helping lift liquids against gravity.

Understanding osmosis not only aids in solving problems about plant physiology but also applies to various disciplines such as chemistry, biology, and environmental science.