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Mice21 [21]
3 years ago
6

A solution of 1.8274g of a polypeptide in 274mL of a aqueous solution has an osmotic pressure at 31.40 degrees C of 2.012 mmHg.

The approximate molecular weight of this polymer is _____ g/mol.
Chemistry
1 answer:
Airida [17]3 years ago
8 0

Answer: Hence, the approximate molecular weight of the poymer is 64106 g/mol

Explanation:

\pi=CRT

where,

\pi = osmotic pressure of the solution = 2.012 mm Hg =  0.0026 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (polypeptide) = 1.8274 g  

Volume of solution = 274 mL

R = Gas constant = 0.0821Latm/Kmol

T = temperature of the solution = 31.40^0C=(273+31.40)K=304.4K

Putting values in above equation, we get:

0.0026=\frac{1.8274 g\times 1000}{M\times 274ml}\times 0.0821\times 304.4

M=64106g/mol

Hence, the approximate molecular weight of the poymer is 64106 g/mol

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2 years ago
How many molecules of H2O and O2 is present in 8.5g of H2O?​
Igoryamba

Answer:

2.8 x 10²³ molecules H₂O

1.4 x 10²³ molecules O₂

Explanation:

First, you will need the balanced chemical equation for the formation of water:

2H₂ + O₂ -> 2H₂O

This will help in determining the mole ratios between water and oxygen, which we will need later.

Let's first calculate the number of H₂O (water) molecules. This will require stoichiometry. We are also given the mass, so we must convert mass into moles, then moles into molecules. mass -> moles -> molecules

8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (6.02 x 10²³ molecules H₂O/1 mol H₂O) = 2.8404 x 10²³ molecules H₂O

Rounded to 2 significant digits: 2.8 x 10²³ molecules H₂O

Now, to find the molecules of water, we can begin with the same stoichiometric equation, but before we convert to molecules, we will have to convert moles of water to moles of oxygen. This is where we will use the mole ratio of water to oxygen we got from the balanced chemical equation earlier. 2H₂O:1O₂

8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (1 mol O₂/2 mol H₂O) x (6.02 x 10²³ molecules O₂/1 mol O₂) = 1.4202 x 10²³ molecules O₂

Rounded to 2 significant digits: 1.4 x 10²³ molecules O₂

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Bonds between carbon and oxygen (c=o) are more polar than bonds between sulfur and oxygen (s=o). nevertheless, sulfur dioxide (s
Harman [31]

Dipole moment can be defined as the product of magnitude of charge and the distance of separation between the charges. A dipole exists when two or more atoms with different electronegativities are bonded together to form a molecule. The resulting unequal sharing of electrons leads to a molecule with a net positive and a net negative end. Hence the molecule is said polar.

Bonds between carbon and oxygen (C=O) are more polar but carbon dioxide (CO2) does not exhibits a dipole moment because CO2 is a linear molecule and the charge is equally distributed amongst the entire molecule. When molecules have an even charge distribution then there is no dipole moment and the molecule is said to be non-polar. CO2 is a linear molecule, so the dipoles are symmetrical and are equal in magnitude but point in opposite directions so they cancel out each other effect and we get net dipole moment zero.

On the other hand sulfur dioxide (SO2) exhibits a dipole moment because unlike CO2 molecule SO2 is not a linear molecule because of the presence of lone pair on Sulfur (S) atom , the geometry of SO2 is bent. This bent orientation of the oxygen's with respect to the sulfur results in the uneven distribution of positive and negative charges between the sulfur atom and the two oxygen atoms in the diagonal-shaped sulfur dioxide molecule. So the dipoles are not equal in magnitude and they do not cancel out and SO2 molecule exhibit a net dipole moment.

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SO2 ( bent shape not linear) = Their is an uneven distribution of positive and negative charges between the sulfur atom and the two oxygen atoms. Dipoles are not equal in magnitude and they do not cancel out and SO2 molecule exhibit a net dipole moment.

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3 years ago
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