Free IB DP Practice Questions | RevisionPrep

1MasterySAQ-SWater potential in plants5 marksPaper 2~8 min

A student places a red blood cell and an Elodea leaf cell into separate beakers of distilled water (

\psi = 0\,\text{kPa}

). The cells are observed under a microscope after 10 minutes.

\text{Red blood cell: } \psi_s = -800\,\text{kPa}

\text{Elodea leaf cell: } \psi_s = -800\,\text{kPa},\quad \psi_p = +600\,\text{kPa}

(a)

Calculate the initial water potential of the Elodea leaf cell before it is placed in distilled water. [1 mark]

(b)

Describe the difference in the appearance of the red blood cell and the Elodea cell after 10 minutes in distilled water. [2 marks]

(c)

Explain, using water potential values, why the Elodea cell reaches equilibrium with the distilled water without bursting, while the red blood cell does not. [2 marks]

Solutions

2MasterySAQ-SWater’s role in temperature regulation5 marksPaper 2~8 min

A coastal city experiences a sea breeze during the day. The land surface heats to

35\,^{\circ}\text{C}

, while the sea surface remains at

22\,^{\circ}\text{C}

. The specific heat capacity of dry soil is

800\,\text{J\,kg}^{-1}\text{K}^{-1}

and the specific heat capacity of seawater is

4200\,\text{J\,kg}^{-1}\text{K}^{-1}

.

(a)

State what is meant by specific heat capacity. [1 mark]

(b)

A

1.0\,\text{kg}

sample of dry soil and a

1.0\,\text{kg}

sample of seawater each absorb

5000\,\text{J}

of solar energy. Calculate the temperature increase of each sample. [2 marks]

(c)

Using your answers from (b), explain how the difference in specific heat capacities between land sea leads to the formation of a sea breeze during the day. [2 marks]

Solutions

3MasterySAQ-SProperties of water5 marksPaper 2~8 min

A marine biologist measures the temperature of ocean water at the surface and at

100\,\text{m}

depth. Surface water is at

25\,°\text{C}

and deep water is at

4\,°\text{C}

. Water reaches its maximum density at

4\,°\text{C}

. Temperature — Density

25\,°\text{C}

—

997\,\text{kg m}^{-3}

4\,°\text{C}

—

1000\,\text{kg m}^{-3}

(a)

State what happens to the density of water as it cools from

25\,°\text{C}

to

4\,°\text{C}

. [1 mark]

(b)

Explain why, in a lake cooling during autumn, water at

4\,°\text{C}

sinks to the bottom while warmer surface water remains above it. [2 marks]

(c)

Discuss how the density properties of water allow aquatic organisms to survive when air temperatures fall well below

0\,°\text{C}

in winter. [2 marks]

Solutions

4MasterySAQ-SProperties of water5 marksPaper 2~8 min

Capillary action contributes to water transport in plants. In a glass tube of radius

0.5\,\text{mm}

, water rises to a height of

5.0\,\text{cm}

. Xylem vessels in a herbaceous plant have an average radius of

0.025\,\text{mm}

.

(a)

State the relationship between the radius of a tube and the height to which water rises by capillary action. [1 mark]

(b)

Calculate the expected height to which water would rise by capillary action alone in the xylem vessels of this plant. [2 marks]

(c)

Evaluate the extent to which capillary action alone can account for water transport to the leaves of a tall tree, and identify one additional mechanism that compensates for any limitation. [2 marks]

Solutions

5MasterySAQ-SProperties of water5 marksPaper 2~8 min

A student investigates osmosis in potato tissue. Cylinders of potato are placed into sucrose solutions of different concentrations. The table shows results after 60 minutes. Sucrose concentration

(\text{mol dm}^{-3})

— Initial mass

(\text{g})

— Final mass

(\text{g})

0.0

—

5.00

—

5.41

0.2

—

5.00

—

5.18

0.4

—

5.00

—

4.97

0.6

—

5.00

—

4.73

0.8

—

5.00

—

4.51

(a)

State the role of hydrogen bonds in the movement of water molecules across a partially permeable membrane. [1 mark]

(b)

Calculate the percentage change in mass of the potato cylinder placed in

0.8\,\text{mol dm}^{-3}

sucrose solution. [2 marks]

(c)

The student claims the water potential of the potato tissue is equivalent to a sucrose concentration of approximately

0.38\,\text{mol dm}^{-3}

. Using the data in the table, evaluate this claim. [2 marks]

Solutions

6MasterySAQ-SProperties of water5 marksPaper 2~8 min

A student places a drop of water on a clean glass surface and observes that it forms a rounded droplet. A second student places a drop of ethanol on the same surface and it spreads into a thin film. - Surface tension of water:

0.073\,\text{N\,m}^{-1}

- Surface tension of ethanol:

0.022\,\text{N\,m}^{-1}

(a)

Describe how hydrogen bonding between water molecules contributes to the rounded shape of the water droplet. [2 marks]

(b)

State one structural difference between water and ethanol molecules that accounts for water's greater capacity to form hydrogen bonds. [1 mark]

(c)

Water striders are insects that move across the surface of ponds. Explain, using the surface tension values given, why water striders can walk on water but would sink on ethanol. [2 marks]

Solutions

7MasterySAQ-SWater potential in plants4 marksPaper 2~6 min

A plant physiologist measures the water potential of a soil sample and finds it to be

-200\,\text{kPa}

. She then measures the water potential of the root cells of a plant growing in the soil and finds it to be

-450\,\text{kPa}

.

(a)

State the direction of water movement between the soil and the root cells. [1 mark]

(b)

Calculate the water potential gradient between the soil and the root cells. [1 mark]

(c)

The plant is watered, and the soil water potential becomes

-80\,\text{kPa}

. The root cells maintain their water potential at

-450\,\text{kPa}

. (i) Calculate the new water potential gradient between the soil and the root cells. [1]
(ii) Explain how the rate of water uptake by the roots would change compared to the original conditions. [2 marks]

Solutions

8MasterySAQ-SProperties of water6 marksPaper 2~9 min

The graph below shows the temperature change over time when equal masses of water and ethanol are heated with identical heaters. [INSERT GRAPH: Two curves. Water curve rises from

20\,^\circ\text{C}

to

35\,^\circ\text{C}

over 10 minutes. Ethanol curve rises from

20\,^\circ\text{C}

to

55\,^\circ\text{C}

over 10 minutes. Axes: Temperature (

^\circ\text{C}

) vs Time (minutes)]

(a)

State the property of water that causes it to heat more slowly than ethanol. [1 mark]

(b)

The specific heat capacity of water is

c_w = 4.18\,\text{J g}^{-1}\,^\circ\text{C}^{-1}

and of ethanol is

c_e = 2.44\,\text{J g}^{-1}\,^\circ\text{C}^{-1}

. Using

Q = mc\Delta T

, calculate the difference in energy absorbed between

50\,\text{g}

of water (temperature change

15\,^\circ\text{C}

) and

50\,\text{g}

of ethanol (temperature change

35\,^\circ\text{C}

) when heated for the same time. [3 marks]

(c)

Explain how the molecular structure of water gives rise to the property stated in (a), and evaluate why this property is important for organisms that use water as a medium for biochemical reactions. [4] Wait — rechecking mark totals: original is 5 marks. (a)[1] + (b)[3] + (c)[4] = 8. Must keep total = 5. Recalibrating: (a)[1] + (b)[2] + (c)[2] = 5, but (c) must be AO3. Keeping (b)[2] and splitting (c) into explanation[2] + evaluation[2] exceeds total. Correct allocation: (a)[1] + (b)[2] + (c)[2] = 5, with (c) rewritten to include a genuine AO3 discriminator within 2 marks. (a) State the property of water that causes it to heat more slowly than ethanol. [1] (b) Using

Q = mc\Delta T

, where

c_w = 4.18\,\text{J g}^{-1}\,^\circ\text{C}^{-1}

and

c_e = 2.44\,\text{J g}^{-1}\,^\circ\text{C}^{-1}

, calculate the energy absorbed by

50\,\text{g}

of water (temperature change

15\,^\circ\text{C}

) and by

50\,\text{g}

of ethanol (temperature change

35\,^\circ\text{C}

). [2] (c) Explain how hydrogen bonding between water molecules gives rise to the property stated in (a), and deduce one consequence of this property for temperature regulation in living organisms. [2 marks]

Solutions

9MasterySAQ-SWater potential in plants5 marksPaper 2~8 min

A plant cell has a solute potential (

\Psi_s

) of

-0.90\,\text{MPa}

and a pressure potential (

\Psi_p

) of

+0.50\,\text{MPa}

. The cell is placed in Solution A, which has a water potential of

-0.30\,\text{MPa}

. A second solution, Solution B, has a solute potential of

-0.65\,\text{MPa}

and a pressure potential of

+0.20\,\text{MPa}

.

(a)

State the water potential of the plant cell at the moment is placed in Solution A. [1 mark]

(b)

Calculate the water potential of Solution B. [2 marks]

(c)

Explain the expected change in the pressure potential of the plant cell over time after it is placed in Solution A. [2 marks]

Solutions

10MasterySAQ-SWater potential in plants5 marksPaper 2~8 min

A biologist studies water relations in a salt-tolerant plant (Spartina alterniflora) growing in a coastal marsh. The root cells have a solute potential (

\Psi_s

) of

-2.4\,\text{MPa}

and a pressure potential (

\Psi_p

) of

+0.6\,\text{MPa}

. The surrounding soil water has a water potential of

-1.2\,\text{MPa}

.

(a)

State the water potential of the root cells. [1 mark]

(b)

The plant is exposed to increased soil salinity, raising the soil water potential to

-2.6\,\text{MPa}

. Calculate the solute potential (

\Psi_s

) the root cells would need to achieve a total water potential of

-3.0\,\text{MPa}

, assuming

\Psi_p

remains

+0.6\,\text{MPa}

. [2 marks]

(c)

Evaluate whether water would move into or out of the root cells under the new conditions described in (b), and explain the physiological significance of the change in

\Psi_s

for Spartina survival in high-salinity environments. [2 marks]

Solutions

11MasterySAQ-SProperties of water10 marksPaper 2~15 min

A gardener notices that on a cold autumn night, a shallow pond freezes over while the deeper part of the same pond remains liquid. The density of ice is

0.917\,\text{g cm}^{-3}

and the density of liquid water at

0\,°\text{C}

is

0.9998\,\text{g cm}^{-3}

.

\rho = \frac{m}{V}

(a)

Describe how the molecular structure of water results in ice being less dense than liquid water. [2 marks]

(b)

Calculate the volume of

100\,\text{g}

of ice. [2 marks]

(c)

Explain how the property of ice floating on water benefits aquatic organisms during winter. [2] Total: [6 marks]

Solutions

12MasterySAQ-SWater’s role in temperature regulation5 marksPaper 2~8 min

A biologist monitors the temperature of a small pond in a temperate forest over 24 hours. The surrounding soil temperature fluctuates by

15\,°\text{C}

during the same period, while the pond temperature fluctuates by only

3\,°\text{C}

. - Surface area of pond:

200\,\text{m}^2

- Average depth of pond:

1.5\,\text{m}

- Density of water:

\rho = 1000\,\text{kg}\,\text{m}^{-3}

- Specific heat capacity of water:

c = 4200\,\text{J}\,\text{kg}^{-1}\,°\text{C}^{-1}

-

Q = mc\Delta T

(a)

State one physical property of water responsible for its resistance to temperature change. [1 mark]

(b)

Explain why the large volume of water in the pond results in a smaller temperature fluctuation than the surrounding soil. [2 marks]

(c)

Calculate thermal energy released by the pond water as it cools by

3.0\,°\text{C}

overnight. Show your working. [2 marks]

Solutions

13MasterySAQ-SWater potential in plants5 marksPaper 2~8 min

A marine biologist studies the water relations of a mangrove tree (Avicennia germinans) growing in a saltwater estuary. The soil water has a solute potential of

-2.5\,\text{MPa}

and a pressure potential of

0\,\text{MPa}

. The root cells of the mangrove have a solute potential of

-3.8\,\text{MPa}

and a pressure potential of

+1.2\,\text{MPa}

. Water potential:

\Psi = \Psi_s + \Psi_p

(a)

State the water potential of the soil water. [1 mark]

(b)

Calculate the water potential of the mangrove root cells. [1 mark]

(c)

Explain whether water will move from the soil into the root cells, and identify the process by which this movement occurs. [3 marks]

Solutions

14MasterySAQ-SProperties of water5 marksPaper 2~8 min

A marine biologist observes that coastal waters cool down more slowly at night than the surrounding land. Water has a specific heat capacity of

4.18\,\text{J g}^{-1}\,°\text{C}^{-1}

, while dry sand has a specific heat capacity of

0.80\,\text{J g}^{-1}\,°\text{C}^{-1}

.

(a)

Explain how hydrogen bonding between water molecules causes water to have a higher specific heat capacity than sand. [1 mark]

(b)

Calculate the energy released when

100\,\text{g}

of water cools from

25\,°\text{C}

to

20\,°\text{C}

. [2 marks]

(c)

Explain one advantage of the slow cooling of coastal water for marine organisms. [2 marks]

Solutions

15MasterySAQ-SWater’s role in temperature regulation7 marksPaper 2~11 min

During exercise, a human athlete produces excess metabolic heat. The body uses evaporative cooling through sweating to regulate temperature. The enthalpy of vaporisation of water is

2.26 \times 10^{6}\,\text{J\,kg}^{-1}

.

(a)

State what is meant by the term latent heat of vaporisation of a substance. [1 mark]

(b)

Explain how the high latent heat of vaporisation of water makes sweating an effective cooling mechanism. [2 marks]

(c)

Calculate the mass of sweat that must evaporate to remove

113\,000\,\text{J}

of heat from the athlete's body. [2 marks]

(d)

The athlete's sweat rate is

1.2\,\text{g\,min}^{-1}

. Not all sweat evaporates; some drips off the skin without evaporating. Evaluate the effectiveness of sweating as a thermoregulatory mechanism under conditions of high humidity, where only

30\%

of sweat produced actually evaporates. [2 marks]

Solutions

16MasterySAQ-SWater as a solvent7 marksPaper 2~11 min

A student investigates the effect of different solvents on the rate of diffusion of a dye through a semi-permeable membrane. Three identical sections of dialysis tubing, each containing

10\,\text{cm}^3

of a

5\%

(mass/volume) solution of blue food dye, are placed into three separate beakers. Beaker 1 contains

100\,\text{cm}^3

of distilled water. Beaker 2 contains

100\,\text{cm}^3

of ethanol (

\text{C}_2\text{H}_5\text{OH}

). Beaker 3 contains

100\,\text{cm}^3

of a

10\%

(mass/volume) sodium chloride (NaCl) solution. The student measures the absorbance of the liquid outside the tubing at

630\,\text{nm}

after 10 minutes. Higher absorbance indicates more dye has diffused out of the tubing. Beaker — Contents of beaker — Absorbance at

630\,\text{nm}

1 — Distilled water —

0.45

2 — Ethanol —

0.08

3 —

10\%

NaCl solution —

0.32

(a)

State the property of water that allows it to dissolve the polar blue food dye. [1 mark]

(b)

Calculate the percentage difference in absorbance between Beaker 1 and Beaker 3, using:

\text{percentage difference} = \frac{\text{absorbance (Beaker 1)} - \text{absorbance (Beaker 3)}}{\text{absorbance (Beaker 3)}} \times 100\%

[2 marks]

(c)

Explain why the absorbance in Beaker 2 is significantly lower than in Beaker 1, with reference to the molecular structures of ethanol and water. [2 marks]

(d)

The student claims that the difference in absorbance between Beaker 1 and Beaker 3 is due solely to osmosis reducing the concentration gradient of dye across the membrane in Beaker 3. Evaluate this claim. [2 marks]

Solutions

17MasterySAQ-SWater as a solvent5 marksPaper 2~8 min

A student investigates the effect of temperature on the rate of starch digestion by the enzyme amylase. Equal volumes of starch solution and amylase solution are mixed at

25\,^\circ\text{C}

and at

50\,^\circ\text{C}

. Iodine solution is added at regular intervals to test for the presence of starch. At

25\,^\circ\text{C}

, the starch is fully digested after 12 minutes; at

50\,^\circ\text{C}

, full digestion occurs after 4 minutes.

(a)

State the relationship between temperature and the rate of starch digestion shown by these results. [1 mark]

(b)

Calculate the percentage decrease in digestion time when temperature is increased from

25\,^\circ\text{C}

to

50\,^\circ\text{C}

. [2 marks]

(c)

Explain, using your knowledge of enzyme structure and molecular kinetics, why the rate of starch digestion is greater at

50\,^\circ\text{C}

than at

25\,^\circ\text{C}

. [2 marks]

Solutions

18MasterySAQ-SWater as a solvent5 marksPaper 2~8 min

Blood plasma is approximately

91\%

water by mass. A patient's blood test shows a plasma calcium concentration of

2.5\,\text{mmol}\,\text{dm}^{-3}

.

(a)

State one property of water that allows it to act as a solvent for ionic compounds. [1 mark]

(b)

Explain why

\text{Ca}^{2+}

ions form stronger interactions with water molecules than

\text{Na}^{+}

ions do. [2 marks]

(c)

Calculate the mass of calcium ions present in

5.0\,\text{dm}^{3}

of this patient's blood plasma. (Molar mass of calcium

= 40.1\,\text{g}\,\text{mol}^{-1}

) [2 marks]

Solutions

19MasterySAQ-SWater’s role in temperature regulation5 marksPaper 2~8 min

A biology field study recorded temperatures of a small pond and surrounding soil over

24

hours. Time — Air temperature / °C — Soil surface temperature / °C — Pond water temperature / °C Midday —

32

—

38

—

26

Midnight —

14

—

9

—

22

(a)

State the property of water that explains why the pond water temperature changed less than the soil temperature over the

24

-hour period. [1 mark]

(b)

Deduce, using data from the table, whether the pond or the soil provides a more stable thermal environment for organisms. [2 marks]

(c)

Evaluate the extent to which thermal property stated in (a) protects aquatic organisms in the pond from the effects of extreme daily temperature fluctuations. [2 marks]

Solutions

20MasterySAQ-SWater’s role in temperature regulation5 marksPaper 2~8 min

In endothermic mammals like the Arctic fox (Vulpes lagopus), evaporative cooling through panting is a key thermoregulatory mechanism. When ambient temperatures exceed

25\,^\circ\text{C}

, the fox pants to lose heat via water evaporation. The latent heat of vaporization of water is

2260\,\text{kJ\,kg}^{-1}

.

(a)

State the property of water that makes evaporative cooling an effective mechanism for heat loss. [1 mark]

(b)

Explain, at the molecular level, how evaporation of water from the respiratory tract removes thermal energy from the fox's body. [2 marks]

(c)

A

5.0\,\text{kg}

Arctic fox loses

0.020\,\text{kg}

of water per hour through panting at

30\,^\circ\text{C}

. Calculate the rate of thermal energy loss, in

\text{kJ\,h}^{-1}

, due to evaporation. [2 marks]

Solutions

Water · Paper 2 — Free Biology SL Practice Questions