Identify each of the following as element, compound, homogenious mixture or heterogenios mixture.

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

Sayid recorded the temperatures of four substances in a chart. which conclusion is best supported by the data?

balandron [24]

The conclusion best supported by the data can either be high temperature, low temperature, and normal temperature. Since there are diverse substances included in the chart, It is expected to also have diverse temperatures.

4 0

3 years ago

In an oven, 7000 kg of limestone are heated to over 900ºC to give lime and carbon dioxide. Assuming all calcium carbonate reacts

kirill [66]

Answer:

3920kg of lime(CaO) 1568m^3 of CO2

Explanation:

I attached the explanation!

also for calculating the volume I assumed Standard Temperature and Pressure(STP).

5 0

3 years ago

Many of water's emergent properties, such as its cohesion, its high specific heat, and its high heat of vaporization, result fro

yarga [219]

Answer:

E) are electrically attracted to each other

Explanation:

Water molecule is polar because there is a difference in electronegativity values between hydrogen and oxygen. The hydrogen side of the molecule has a slight positive charge and the oxygen side is slightly negatively.

Positively and negatively charged ends cause water molecules to attract one another and for this reason water shows the properties mentioned in the question: cohesion, high specific heat, and high heat of vaporization.

3 0

3 years ago

"A solution with a total volume of 850.0 mL contains 43.5 g Mg(NO3)2. If you remove 25.0 mL of this solution and then dilute thi

ycow [4]

Answer: C = 0.014M

Explanation:

From n= m/M= CV

m =43.5 M= 148, V=850ml

43.5/148= C× 0.85

C= 0.35M

Applying dilution formula

C1V1=C2V2

C1= 0.35, V1= 25ml, C2=?, V2= 600ml

0.35× 25 = C2× 600

C2= 0.014M

7 0

3 years ago