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3 years ago

Convert 1.39 x 10^24 atoms of carbon to moles of carbon

2 answers:

8 0

If 1mol ------- is ----------- 6,02×10²³
so x ------- is ----------- 1,39×10²⁴

$x=\frac{1,39*10^{24}*1mol}{6,02*10^{23}}\approx2,31mol$

Delicious77 [7]3 years ago

7 0

Answer is: there are 2.308 moles of carbon.

N(C) = 1.39·10²⁴; number of carbon atoms.

n(C) = N(C) ÷ Na.

n(C) = 1.39·10²⁴ ÷ 6.022·10²³ 1/mol.

n(C) = 2.308 mol; amount of carbon atoms.

Na is Avogadro number or Avogadro constant (the number of particles, in this example carbon, that are contained in the amount of substance given by one mole).

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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

sammy [17]

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,

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Putting values in above equation, we get:

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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,

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

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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

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