Part A 1.50 L of 0.130 M (NH4)2SO4 solution, starting with solid (NH4)2SO4 Enter

Question

Part A

1.50 L of 0.130 M (NH4)2SO4 solution, starting with solid (NH4)2SO4

Enter your answers numerically separated by a comma.

25.8,1.50

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Correct

Part B

230 g of a solution that is 0.65 m in Na2CO3, starting with the solid solute

Enter your answers numerically separated by a comma.

1068.5,121.6

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Incorrect; Try Again; no points deducted

Part C

1.30 L of a solution that is 16.0 % of Pb(NO3)2 by mass (the density of the solution is 1.16 g/mL), starting with solid solute

Enter your answers numerically separated by a comma.

86.21,231.27

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Incorrect; Try Again

Part D

a 0.50M solution of HCl that would just neutralize 4.5 g of Ba(OH)2 starting with 6.0 M HCl

Express your answer using two significant figures. Enter your answers numerically separated by a comma.

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25.8,1.50

Explanation / Answer

Solution :-

Part A

1.50 L of 0.130 M (NH4)2SO4 solution, starting with solid (NH4)2SO4

Solution :- Letscalculaet the moles of the (NH4)2SO4

Moles = molarity * volume in liter

           = 0.130 mol per L * 1.50 L

           = 0.195 mol (NH4)2SO4

Now lets calculate its mass

Mass of (NH4)2SO4 = moles * molar mass

                                   = 0.195 mol * 132.14 g per mol

                                   = 25.8 g (NH4)2SO4

So the answer is 25.8 g (NH4)2SO4 and volume of water is 1.50 L

Part B

230 g of a solution that is 0.65 m in Na2CO3, starting with the solid solute

Solution :- Lets calculate the moles of the Na2CO3

Lets calculate the mass of the Na2CO3

Suppose we have 1kg solvent

Then moles of Na2CO3 = 0.65 moles

0.65 mol Na2CO3 * 105.9888 g per mol = 68.89 g

So total mass of solution = 1000 g + 68.69 g = 1068.69 g

So lets calculate the mass of the Na2CO3 in 230 g solution

230 g * 68.69 g Na2CO3 /1068.69 g = 14.78 g

So mass of water = 230 g – 14.78 g = 215.22 g

So mass of Na2CO3 = 14.78 g and mass of water = 215.22 g

Part C

1.30 L of a solution that is 16.0 % of Pb(NO3)2 by mass (the density of the solution is 1.16 g/mL), starting with solid solute

Solution :-

Lets calculate the mass of the solution

Mass of solution = volume * density

                              = (1.30 L*1000 ml /1 L)*(1.16 g / 1 ml )

                              = 1508 g

Now lets calculate the mass of Pb(NO3)2 using the percentage

1508 g solution * 16 % Pb(NO3)2 / 100 % = 241.08 g

Mass of water = 1508 g – 241.08 g = 1266.72 g

So mass of Pb(NO3)2 = 241.08 g and mass of water = 1266.72 g

Part D

a 0.50M solution of HCl that would just neutralize 4.5 g of Ba(OH)2 starting with 6.0 M HCl

Express your answer using two significant figures.

Solution :- lets calculate moles of Ba(OH)2

Moles of Ba(OH)2 = 4.5 g / 171.34 g per mol = 0.02626 mol

Moles of HCl = 0.02626 mol Ba(OH)2 * 2 mol HCl / 1 mol Ba(OH)2 = 0.05252 mol HCl

Now lets calculate the volume of the 6.0 M HCl

Volume = moles / molarity

              = 0.05252 mol / 6.0 mol per L

              = 0.00875 L

0.00875 L * 1000 ml / 1 L = 8.75 L

So we need 8.75 ml of 6.0 M HCl

Now lets calculate the volume of the 0.50 M HCl

M1V1*M2V2

V2= M1V1/M2

V2= 6.0 M * 8.75 ml / 0.50 M

V2 = 105 ml

So answer is

8.75 ml 6.0 M HCl and 105 ml of 0.50 M HCl

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