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3 years ago

5

In a solution there is usually more __than _ A) solute/solvent B) soluble/insoluble C) saturation/unsaturati

on D) solvent/solute

1 answer:

Elan Coil [88]3 years ago

5 0

Answer:

I think the answer is option 3

You might be interested in

A potassium atom can achieve a stable octet by

nydimaria [60]

C. losing one electron

Explanation:

It is because the potassium atom electronic configuration is 2,8,8,1 where by if it loses one electron it becomes stable

6 0

3 years ago

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

HELP ME PLS I NEED HELP

sesenic [268]

Lol your answer is c

3 0

3 years ago

How do the densities of the two objects compare?

Marizza181 [45]

Compare the density of the object in question to the density of water. If its density is less than water, it will float. For example, oak floats because its density is 0.7 g/cm³. If the density of an object is greater than water, it will sink.

3 0

3 years ago

2c4H10 +13O2 —>8CO2+10H2O to find how many moles of oxygen would react with 4.3 mol C4H10

Elena-2011 [213]

Answer:

$\large\boxed{\text{28 mol of O$_{2}$}}$

Explanation:

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

n/mol: 4.3

13 mol of O₂ react with 2 mol of 2C₄H₁₀

$\text{Moles of O}_{2} = \text{4.3 mol C$_{4}$H$_{10}$} \times \dfrac{\text{13 mol O}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \textbf{28 mol O}_{2}\\\\\text{The reaction requires $\large\boxed{\textbf{28 mol of O$_{2}$}}$}$

5 0

3 years ago