Two properties of non-metallic elements and their

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

The maximum allowable concentration of pb2+ ions in drinking water is 0.05 ppm (i.e., 0.05 g of pb2+ in 1 million grams of water

trasher [3.6K]

PbSO₄ partially dissociates in water. the balanced equation is;

PbSO₄(s) ⇄ Pb²⁺(aq) + SO₄²⁻(aq)
Initial - -
Change -X +X +X
Equilibrium X X

Ksp = [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸ = X * X
1.6 x 10⁻⁸ = X²
X = 1.3 x 10⁻⁴ M

Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M.
[Pb²⁺] = 1.3 x 10⁻⁴ M.
= 1.3 x 10⁻⁴ mol / L x 207 g / mol
= 26.91 ppm

8 0

3 years ago

How many grams are in 9.05 x 1023 atoms of silicon

UkoKoshka [18]

Answer:

42.2075 grams

Explanation:

4 0

2 years ago

PLEASE ANSWER I WILL GIVE YOU BRAINIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

blondinia [14]

Answer:

Erosion

Explanation:

Erosion is not a cause of metamorphism

The causes are mostly temperature, pressure, heat from the metamorphic rock

5 0

2 years ago

Read 2 more answers

(Thermodynamics)

frutty [35]

Answer:

3853 g

Step-by-step explanation:

M_r: 107.87

16Ag + S₈ ⟶ 8Ag₂S; ΔH°f = -31.8 kJ·mol⁻¹

1. Calculate the moles of Ag₂S

Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S

2. Calculate the moles of Ag

Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag

3. Calculate the mass of Ag

Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag

You must react 3853 g of Ag to produce 567.9 kJ of heat

3 0

3 years ago