Respuesta :
Answer:
211.4g/mol.
Explanation:
Data obtained from the question includes:
Mass of unknown compound = 0.98g
Mass of solvent = 10.30g
Molality = 0.45 M
Next, we shall determine the number of mole of the unknown compound present in the solution.
This can be obtained as follow:
Molality = mole /kg of solvent
Mole of the unknown compound =.?
Mass of solvent = 10.30g = 10.30/1000 = 0.0103Kg
Molality = 0.45 M
Molality = mole /kg of solvent
0.45 = mole /0.0103
Cross multiply
Mole = 0.45 x 0.0103
Mole = 4.635×10¯³ mole
Therefore the mole of the unknown compound that dissolve in solution is 4.635×10¯³ mole
Now, we can obtain the molar mass of the unknown compound as follow:
Mole of the unknown compound = 4.635×10¯³ mole
Mass of unknown compound = 0.98g
Molar mass of the unknown compound =?
Mole = mass /Molar mass
4.635×10¯³ = 0.98 /Molar mass
Cross multiply
4.635×10¯³ x molar mass = 0.98
Divide both side by 4.635×10¯³
Molar mass = 0.98 / 4.635×10¯³
Molar mass = 211.4g/mol.
Therefore, the molar mass of the unknown compound is 211.4g/mol.
The molecular mass of the unknown has been 211.66 g/mol.
Molality can be defined as the moles of the solute per kg of solvent.
Molality can be expressed as:
Molality = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{molecular\;mass\;of\;solute}\;\times\;\dfrac{1000}{Mass\;of\;solvent\;(g)}[/tex] ......(i)
The given unknown has been the solute.
The mass of solute = 0.98 g.
The mass of solvent = 10.30 g.
The molality of the solution formed has been = 0.45 m.
Substituting the values in equation (i):
0.45 m = [tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}\;\times\;\dfrac{1000}{10.30\;g}[/tex]
0.45 m = [tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}\;\times\;97.087[/tex]
[tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}[/tex] = [tex]\rm \dfrac{0.45}{97.087}[/tex]
[tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}[/tex] = 0.00463
Molecular mass of solute = [tex]\rm \dfrac{0.98}{0.00463}[/tex]
Molecular mass of solute = 211.66 g/mol.
The molecular mass of the unknown has been 211.66 g/mol.
For more information about the molality of the solution, refer to the link:
https://brainly.com/question/8103026
