In: Chemistry
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.
m(NH4)2SO4,VH2O = |
25.8,1.50 |
g,L |
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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.
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||||
mNa2CO3,mH2O = |
1068.5,121.6 |
g,g |
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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.
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||||
mPb(NO3)2,mH2O = |
86.21,231.27 |
g,g |
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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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||||
V6MHCl,V0.5MHCl = | mL,mL |
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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