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forsale [732]
3 years ago
7

A 2.00-g sample of a large biomolecule was dissolved in 15.0 g carbon tetrachloride. the boiling point of this solution was dete

rmined to be 77.85c. calculate the molar mass of the biomolecule. for carbon tetrachloride, the boiling-point constant is 5.03c kg/mol, and the boiling point of pure carbon tetrachloride is 76.50c.
Chemistry
1 answer:
Katarina [22]3 years ago
5 0
We will use boiling point formula:

ΔT = i Kb m 

when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35

and Kb is the boiling point constant =5.03

and m = molality 

i = vant's Hoff factor

so by substitution, we can get the molality:

1.35 = 1 * 5.03 * m

∴ m = 0.27

when molality = moles / mass  Kg

           0.27 = moles /  0.015Kg

∴ moles = 0.00405 moles

∴ The molar mass = mass / moles
                               = 2 g /  0.00405 moles 
                               = 493.8 g /mol
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A 20.0 gram sample of an element contains 4.95 x 1023 atoms. what is the element?
Sonja [21]
You will want to find how many grams are in a whole mole so you know which element it is. To do this, find out how much of a mole you have.

4.95 x 10^23 atoms / 6.022 x 10^23 atoms (one whole mole of any element) = .8219860511 or ~82% of 1 mole

Now we know that, find what to multiply 20 g by to get the rest of the mole.

1 mole / .8219860511 mole = 1.216565657

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Now that you have grams per mole, you can look at the periodic table and the molar masses to see which this number is closely aligned.

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2 years ago
How many moles of argon are contained in 58 L of At at STP?
Harman [31]

Answer:

n = 2.58 mol

Explanation:

Given data:

Number of moles of argon = ?

Volume occupy = 58 L

Temperature = 273.15 K

Pressure = 1 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

1 atm × 58 L = n × 0.0821 atm.L/ mol.K × 273.15 K

58 atm.L = n × 22.43 atm.L/ mol.

n = 58 atm.L / 22.43 atm.L/ mol

n = 2.58 mol

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