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VMariaS [17]
3 years ago
7

Sean plated an unknown metal onto his silver earring which initially weighed 1.41 g. He constructs an electrolytic cell using hi

s ring as one of the electrodes. After running the cell, 0.022 moles of the unknown metal was plated onto his ring and the mass of the ring increased to 3.952 g. What is the atomic weight of the unknown metal in g/mol
Chemistry
1 answer:
goldenfox [79]3 years ago
3 0

Answer:

115 g/mol

Explanation:

Step 1: Given data

  • Initial mass of the earring (mi): 1.41 g
  • Final mass of the earring (mf): 3.952 g
  • Moles of the unknown metal deposited (n): 0.022 mol

Step 2: Calculate the mass of the unknown metal deposited

An earring is plated in an electrolytic cell. The mass of the unknown metal deposited can be calculated using the following expression.

<em>m = mf - mi</em>

m = 3.953 g - 1.41 g

m = 2.54 g

Step 3: Calculate the molar mass (M) of the unknown metal

0.022 moles of the metal have a mass of 2.54 g. The molar mass of the metal is:

<em>M = m/n</em>

M = 2.54 g / 0.022 mol

M = 115 g/mol

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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.

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