1. What is the mass of sulfuric acid, H2SO4, in 25.0 mL of 12.0 M sulfuric acid?

Question

1. What is the mass of sulfuric acid, H2SO4, in 25.0 mL of 12.0 M sulfuric acid?

2. If 25.0 mL of 0.150 M aluminum bromide reacts with 15.0 mL of 0.200 M silver nitrate, what mass of silver bromide is produced? _AlBr3 (aq) +_AgNO3 (aq) -> _AgBr (s) + _Al(NO3)3 (aq)

3. What is the molarity of aluminum ions in the above question before the reaction has taken place? (Above Question 2)

4. Calculate the boiling point of a 0.4500 m solution of ethylene glycol.

5. What is the pH of a 5.40 x 10^-2 M barium hydroxide solution?

Explanation / Answer

1. What is the mass of sulfuric acid, H2SO4, in 25.0 mL of 12.0 M sulfuric acid?

Solution :- Lets first calculate the moles of H2SO4

Moles of H2SO4 = molarity * volume in liter

                            = 12.0 mol per L * 0.025 L

                           = 0.300 mol

Mass of H2SO4 = moles * molar mass

                           =0.300 mol * 98.08 g per mol

                           = 29.4 g H2SO4

2. If 25.0 mL of 0.150 M aluminum bromide reacts with 15.0 mL of 0.200 M silver nitrate, what mass of silver bromide is produced? _

Solution :-

Balanced reaction equation

AlBr3 (aq) +3AgNO3 (aq) -> 3AgBr (s) + Al(NO3)3 (aq)

Lets calculate the moles of each reactant

Moles of AlBr3 = 0.150 mol per L * 0.025 L = 0.00375 mol

Moles of AgNO3 = 0.200 mol per L * 0.015 L = 0.003 mol AgNO3

Moles of AgNO3 are less than moles of AlBr3

And mole ratio is 1 mol AlBr3 = 3 mol AgNO3

Therefore AgNO3 is the limiting reactant

So the moles of AgBr that can be formed are calculated

0.003 mol AgNO3 * 3 mol AgBr / 3 mol AgNO3 = 0.003 mol AgBr

Now lets convert moles of AgBr to its mass

Mass of AgBr = moles * molar mass   

                       = 0.003 mol * 187.77 g per mol

                        = 0.563 g AgBr

3. What is the molarity of aluminum ions in the above question before the reaction has taken place? (Above Question 2)

Solution :- AlBr3 concetration = 0.150 M so the molarity of the Al^3+ = 0.150 M in the original solution 25.0 ml

When they mixed then volume changes to 25 +15 = 40 ml

So the concentration of the Al^3+ ion = 0.150 M * 25.0 ml / 40.0 ml = 0.0938 M

4. Calculate the boiling point of a 0.4500 m solution of ethylene glycol.

Solution :-

Delta Tb = Kb * m

Kb of water is 0.512 C/m

Delta Tb = 0.512 C per m * 0.4500 m

Delta Tb = 0.2304 C

So the boiling point of solution is

BP of solution = 100 C + 0.2304 C = 100.23 C

5. What is the pH of a 5.40 x 10^-2 M barium hydroxide solution?

Solution :- Ba(OH)2 ----- > Ba^2+ + 2 OH-

So the concentration of the OH- = 5.40*10^-2 M * 2 = 1.08*10^-1 M

pOH= -log [OH-]

pOH = -log [1.08*10^-1]

pOH= 0.97

pH + pOH= 14

pH= 14 – pOH

pH= 14 – 0.97

pH = 13.03

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.