ILL GIVE U BRAINILIEST

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

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

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

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

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

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

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

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

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

2 years ago

If 450.0 mL of a 0.500 M solution is mixed with 200.0 mL of water, what is the molarity of the new

Oksana_A [137]

Answer:

Answer: A) .346 M

Explanation:

Given:

- 450 mL

- .5 M soln

-200 mL water

1) Convert mL to L

450 mL = .45 L

200 mL = .2 L

2) Find mols in solution

.5 M = x/.45 L

x = .225 mol

3) Find total volume of solution

.45 L + .2 L =.65 L

4) Find new molarity

molarity (M) = mols solute/ L solution

y = .225 mol (from step 2)/ .65 L (from step 3)

y = .346 M

Answer: A) .346 M

7 0

2 years ago

Arrange the following elements in order of increasing ionization energy

Gnoma [55]

Ionization Trend: First ionization energy will increase left to right across a period and increase bottom to top of a family (column).

A) Sr, Be, Mg are all in column 2 of the periodic table. Based on the first ionization rule above, from increasing to decreasing energy, the order is: Be, Mg, Sr

B) Bi, Cs, Ba are all in the same row of the periodic table. Based on the first ionization rule above, from increasing to decreasing energy, the order is: Bi, Ba, Cs

C) Same rule as above. Order is: Na, Al, S

8 0

2 years ago

How many moles of carbon are there in a 0.70 g (3.5 carat) diamond?

maksim [4K]

Answer:

3.5 x 10^22 moles of carbon are there in a 0.70 g (3.5 carat) diamond.

Explanation:

In case of diamond, the number of atoms is the same as the number of molecules.

7 0

3 years ago

A valid lewis structure of __________ cannot be drawn without violating the octet rule. a valid lewis structure of __________ ca

Oxana [17]

A valid Lewis structure of IF3 cannot be drawn without violating the octet rule.

Answer: IF3 (Iodine Trifluoride)

This is because, I (Iodine) and F (Fluorine) both have odd number of valence electrons (7) which also means that there are too many valence electrons in the valence shell.

5 0

3 years ago