A mug is filled with coffee. Which property is an intensive property of the coffee in the mugg

At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

8 0

3 years ago

A student made the following diagram to represent cellular respiration.

Ratling [72]

The correct answer is A, Water is not used up during this process. This is because when cellular respiration occurs oxygen and glucose combine. When this takes place water is left behind when carbon is separated from glucose. Because water is being left behind it is not being used up in this process.

3 0

3 years ago

Read 2 more answers

Which of these properties is the best one to use for identification of an element? 1.the number of neutrons in the atomic nucleu

Verdich [7]

The right answer is 2.

The number of protons contained in a nucleus (called an atomic number) is characteristic of a chemical element. For a given atomic number, the number of neutrons defines different "types" of this element: isotopes. The variation of the number of protons of the nucleus of an atom, during a nuclear reaction for example, causes a change of the element studied.

4 0

3 years ago

Read 2 more answers

Please help me. We dally transform energy in our homes and schools into light, sound, heat, etc. what is the most common energy

Irina-Kira [14]

I hope this helps sorry if it doesn’t

7 0

3 years ago

An aluminum block has a density of 2.70 g/mL. If the mass of the block is 24.60 g, find the volume of the substance.

harina [27]

Volume of a substance can be determined by dividing mass of the substance by its density.

That can be mathematical shown as:

Density=Mass/Volume

So, Volume=Mass/Density

Here mass of the substance given as 24.60 g

Whereas density of the substance is 2.70 g/mL

So,

Volume=Mass/Density

=24.6/2.7

=9.1 mL

So volume of the substance is 9.1 mL.

8 0

3 years ago