Chemistry, 02.10.2019 22:00 kenoknox

Asolution was prepared by dissolving 0.800 g of sulfur s8, in 100.0 g of acetic acid, hc2h3o2. calculate the freezing point and boiling point of the solution . for the solvent, kb = 3.08 oc/m and kf = 3.59 oc/m.

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Answer from: skoolroks

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

Answer from: Quest

Sorry i don’t know i just need the points

Answer from: Quest

the grams of barium that are in 2.3 x 10^23 atoms of barium is 52.4 g (answer c)

calculation

mass = moles of ba x molar mass of ba

molar mass of ba from periodic table= 137.3 g/mol

use the avogadro's law constant to calculate the moles of barium

that is 1 mole = 6.02 x 10^23 atoms

? moles = 2.3 x 10^23 atoms

by cross multiplication

=(1 mole x 2.3x 10^23 atoms) / 6.02 x10^23 atoms)= 0.382 moles

mass is therefore = 137.3 g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

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Asolution was prepared by dissolving 0.800 g of sulfur s8, in 100.0 g of acetic acid, hc2h3o2. calcu...

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