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Debora [2.8K]
3 years ago
10

Calculate the number of moles and the mass of the solute in each of the following solutions:

Chemistry
1 answer:
frosja888 [35]3 years ago
3 0

<u>Answer:</u>

<u>For a:</u> The number of moles of KI are 2.7\times 10^{-5} and mass is 4.482\times 10^{-3}g

<u>For b:</u> The number of moles of sulfuric acid are 1.65\times 10^{-5} and mass is 1.617\times 10^{-3}g

<u>For c:</u> The number of moles of potassium chromate are 2.84\times 10^{-2} and mass is 5.51 g.

<u>For d:</u> The number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}    .....(1)

To calculate the number of moles of a substance, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(2)

  • <u>For a:</u>

Molarity of KI = 8.23\times 10^{-5}M

Volume of solution = 325 mL = 0.325 L     (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

8.25\times 10^{-5}mol/L=\frac{\text{Moles of KI}}{0.325L}\\\\\text{Moles of KI}=2.7\times 10^{-5}mol

Now, using equation 2, we get:

Moles of KI = 2.7\times 10^{-5}mol

Molar mass of KI = 166 g/mol

Putting values in equation 2, we get:

2.7\times 10^{-5}mol=\frac{\text{Mass of KI}}{166g/mol}\\\\\text{Mass of KI}=4.482\times 10^{-3}g

Hence, the number of moles of KI are 2.7\times 10^{-5} and mass is 4.482\times 10^{-3}g

  • <u>For b:</u>

Molarity of sulfuric acid = 22\times 10^{-5}M

Volume of solution = 75 mL = 0.075 L

Putting values in equation 1, we get:

22\times 10^{-5}mol/L=\frac{\text{Moles of sulfuric acid}}{0.075L}\\\\\text{Moles of }H_2SO_4=1.65\times 10^{-5}mol

Now, using equation 2, we get:

Moles of sulfuric acid = 1.65\times 10^{-5}mol

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 2, we get:

1.65\times 10^{-5}mol=\frac{\text{Mass of }H_2SO_4}{98g/mol}\\\\\text{Mass of }H_2SO_4=1.617\times 10^{-3}g

Hence, the number of moles of sulfuric acid are 1.65\times 10^{-5} and mass is 1.617\times 10^{-3}g

  • <u>For c:</u>

Molarity of potassium chromate = 0.1135M

Volume of solution = 0.250 L

Putting values in equation 1, we get:

0.1135mol/L=\frac{\text{Moles of }K_2CrO_4}{0.250L}\\\\\text{Moles of }K_2CrO_4=2.84\times 10^{-2}mol

Now, using equation 2, we get:

Moles of potassium chromate = 2.84\times 10^{-2}mol

Molar mass of potassium chromate = 194.2 g/mol

Putting values in equation 2, we get:

2.84\times 10^{-2}mol=\frac{\text{Mass of }K_2CrO_4}{194.2g/mol}\\\\\text{Mass of }K_2CrO_4=5.51g

Hence, the number of moles of potassium chromate are 2.84\times 10^{-2} and mass is 5.51 g.

  • <u>For d:</u>

Molarity of ammonium sulfate = 3.716 M

Volume of solution = 10.5 L

Putting values in equation 1, we get:

3.716mol/L=\frac{\text{Moles of }(NH_4)_2SO_4}{10.5L}\\\\\text{Moles of }(NH_4)_2SO_4=39.018mol

Now, using equation 2, we get:

Moles of ammonium sulfate = 39.018 mol

Molar mass of ammonium sulfate = 132.14 g/mol

Putting values in equation 2, we get:

39.018mol=\frac{\text{Mass of }(NH_4)_2SO_4}{132.14g/mol}\\\\\text{Mass of }(NH_4)_2SO_4=5155.84g

Hence, the number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

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Answer:

About 67 grams or 67.39 grams

Explanation:

First you would have to remember a few things:

 enthalpy to melt ice is called enthalpy of fusion.  this value is 6.02kJ/mol

  of ice  

 it takes 4.18 joules to raise 1 gram of liquid water 1 degree C

 water boils at 100 degrees C and water melts above 0 degrees C

 1 kilojoules is 1000 joules

  water's enthalpy of vaporization (steam) is 40.68 kJ/mol

  a mole of water is 18.02 grams

  we also have to assume the ice is at 0 degrees C

Step 1

Now start with your ice.  The enthalpy of fusion for ice is calculated with this formula:

q = n x ΔH    q= energy, n = moles of water, ΔH=enthalpy of fusion

Calculate how many moles of ice you have:

150g x (1 mol / 18.02 g) = 8.32 moles

Put that into the equation:

q = 8.32 mol x 6.02 = 50.09 kJ of energy to melt 150g of ice

Step 2

To raise 1 gram of water to the boiling point, it would take 4.18 joules times 100 (degrees C)  or 418 joules.

So if it takes 418 joules for just 1 gram of water, it would take 150 times that amount to raise 150g to 100 degrees C.  418 x 150 = 62,700 joules or 62.7 kilojoules.

So far you have already used 50.09 kJ to melt the ice and another 62.7 kJ to bring the water to boiling.  That's a total of 112.79 kJ.

Step 3

The final step is to see how much energy is left to vaporize the water.

Subtract the energy you used so far from what you were told you have.

265 kJ - 112.79 kJ = 152.21 kJ

Again q = mol x ΔH (vaporization)

You know you only have 152.21 kJ left so find out how many moles that will vaporize.

152.21 kJ = mol x 40.68  or   mol = 152.21 / 40.68  = 3.74 moles

This tells you that you have vaporized 3.74 moles with the energy you have left.

Convert that back to grams.

3.74 mol   x  ( 18.02 g / 1 mol ) = 67.39 grams

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