Which two notations represent isotopes of the same element?

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

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. Cac

Tanya [424]

Answer:

The equation that gives the overall equilibrium in terms of the equilibrium constants K and Ky is K1 = K^6 * Ky

Explanation:

we have the following balanced reaction:

CaC2 + 2H2O = C2H2 + Ca(OH)2

the value of K for this reaction will be equal to:

K = ([C2H2] * [Ca(OH)2])/([CaC2] * [H2O]^2)

if we multiply the reaction by the value of 6, we have:

6CaC2 + 12H2O = 6C2H2 + 6Ca(OH)2

Again, the value of K for this reaction will be equal to:

K,´ = ([C2H2] ^6 * [Ca(OH)2]^6)/([CaC2]^6 * [H2O]^12) = K^6

For the second reaction:

6C2H2 + 3CO2 + 4H2O = 5CH2CHCO2H

The value of K for this reaction:

K2 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^4)

we also have:

K1 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^16)

Thus:

K1 = K^6 * Ky

4 0

2 years ago

2. Al + Fe2O3 → Al2O3 + Fe (Need to balance equation)

shtirl [24]

Balanced equation: 2Al+Fe2O3–> Al2O3+2Fe

Using mole calculations you can find that 10g of Al produces 24.3g of Fe under the conditions described

4 0

2 years ago

Will marks brainly if it isn’t a copied answer!!!

Neporo4naja [7]

What kind of control like scientific?

5 0

2 years ago

Read 2 more answers

Explain the three laws of chemical reaction in detain gilling examples

kotykmax [81]

Answer:

The three laws of Chemical Reaction are .The law of constant proportions. The law of multiple proportions. The law of reciprocal proportions.

A chemical compound is always found to be made up of the same elements combined together in the same fixed proportion by mass.

potassium and chlorine gas ---> chloride.

Hope this helps, have a good day.✌

8 0

2 years ago