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elena-14-01-66 [18.8K]
3 years ago
10

Using the periodic table determine the atomic mass of Na2C2O4

Chemistry
1 answer:
Mkey [24]3 years ago
4 0
Masses of atoms are the sum of neutrons and protons. Atomic mass given for the element is the weighted atomic masses of the isotopes depending on the abundance of the isotopes.
Atomic masses of the elements making up Na₂C₂O₄ are as follows;
Na - 22.98 a.m.u
2 atoms of Na - 22.98 x 2 = 45.96
C - 12.01 a.m.u
2 atoms of C - 12.01 x 2 = 24.02
O - 15.99 a.m.u
4 atoms of O - 15.99 x 4 = 63.96
Sum of the atomic masses = 45.96 + 24.02 + 63.96 = 133.94 
Mass is 133.94 g/mol
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Answer:

The correct option is C.

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The statement given in option C about glucose is wrong because glucose is a monosaccharide and not a disaccharide.

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Groups of elements that are shiny but not lustrous, have a semi-conductive property, and are brittle are classified as?
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A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an
inysia [295]

<u>Answer:</u> The molar mass of the unknown compound is 223.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

where,

\pi = osmotic pressure of the solution = 8.44 torr

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

M = molarity of solute = ?

R = Gas constant = 62.3637\text{ L torr }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

8.44torr=1\times M\times 62.3637\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\M=\frac{8.44}{1\times 62.3637\times 298}=4.54\times 10^{-4}M

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 4.54\times 10^{-4}M

Given mass of unknown compound = 15.2 mg = 0.0152 g   (Conversion factor:  1 g = 1000 mg)

Volume of solution = 150.0 mL

Putting values in above equation, we get:

4.54\times 10^{-4}M=\frac{0.0152\times 1000}{\text{Molar mass of unknown compound}\times 150.0}\\\\\text{Molar mass of unknown compound}=\frac{0.0152\times 1000}{150.0\times 4.54\times 10^{-4}}=223.2g/mol

Hence, the molar mass of the unknown compound is 223.2 g/mol

5 0
3 years ago
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