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musickatia [10]
3 years ago
9

Determine the overall charge on each complex.a) tetrachlorocuprate(i)b) pentaamminechlorocobalt(iii)c) diaquadichloroethylenedia

minecobalt(iii)
Chemistry
1 answer:
balu736 [363]3 years ago
4 0
Complex a: <span>tetrachlorocuprate(i)
In present complex chlorine is negatively charged ligand (-1) and oxidation state of copper is +1. Therefore, total charge on complex = 4(-1) + 1 = -3

Complex b: </span><span>pentaamminechlorocobalt(iii)
In present complex, ammine is a neutral ligand (charge = 0), chlorine is negatively charged ligand (charge = -1) and oxidation state of Co is +3. Therefore, total charge on complex is 5(0) + (-1) + (+3) = +2

</span><span>Complex c: diaquadichloroethylenediaminecobalt(iii)
</span>In present complex, aqua is a neutral ligand (charge = 0), chlorine is a negatively charged ligand (charge = -1), ethylenediamine is a neutral ligand (charge = 0) and oxidation state of cobalt is +3. Therefore, total charge on complex is 2(0) + 2(-1) + 2(0) + 3 = +1. 
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The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa
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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 MgCl_2 (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 MgCl_2 (solute) = 3.37 g

First we have to calculate the mass of solute.

\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2

\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g

Now we have to calculate the mass of solution.

\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g

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.

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

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%

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.

8 0
3 years ago
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