Respuesta :
Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.
Solution : Given,
Density of solution = 1.25 g/ml
Molar mass of [tex]MgCl_2[/tex] (solute) = 95.21 g/mole
3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.
The volume of solution = 1 L = 1000 ml
Mass of [tex]MgCl_2[/tex] (solute) = 3.37 g
First we have to calculate the mass of solute.
[tex]\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2[/tex]
[tex]\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g[/tex]
Now we have to calculate the mass of solution.
[tex]\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g[/tex]
Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g
Now we have to calculate the molality of the solution.
[tex]Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg[/tex]
The molality of the solution is, 0.0381 mole/Kg.
Now we have to calculate the mass/mass percent.
[tex]\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%[/tex]
The mass/mass percent is, 25.67 %
Now we have to calculate the mass/volume percent.
[tex]\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%[/tex]
The mass/volume percent is, 32.086 %
Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.
