Question 10
The following were the titration results of 25.0 cm3 of a sodium hydroxide solution
with 0.10M nitric acid:
Burette readings (cm3) 1st 2nd 3rd
Final reading 32.5 32.5 33.2
Initial reading 1.0 2.6 3.1
The molarity of the sodium hydroxide solution is
A. 0.120M
B. 0.122M
C. 0.124M
D. 0.126M
Question 11
The following is the result of the titration of 25.00 cm3 of sodium hydroxide solution
with 1.00M hydrochloric acid:
Burette readings (cm3) 1st 2nd 3rd
Final reading 21.60 21.60 21.80
Initial reading 1.00 1.50 1.90
The concentration of the sodium hydroxide solution is
A. 0.796M
B. 0.800M
C. 0.804M
D. 0.808M
Question 12
1.43g of hydrated sodium carbonate Na2CO3.nH2O is dissolved in water and the
resulting solution is made up to 200 cm3. 25 cm3 of the solution requires 12.5 cm3 of
0.1M hydrochloric acid for complete reaction. The value of n is
A. 1
B. 5
C. 7
D. 10
Question 13
Each of the following 0.50M acids is titrated with 0.5M sodium hydroxide solution.
Which sample requires the greatest volume of sodium hydroxide for complete
neutralization?
A. 40.0 cm3 of H3PO4 (a tribasic acid)
B. 45.0 cm3 of H2SO3 (a dibasic acid)
C. 45.0 cm3 of H2SO4
D. 90.0 cm3 of HNO3
4
Question 14
The volume of 0.1M sodium carbonate which will completely react with 25 cm3 of
0.2M phosphoric acid (H3PO4) to give a solution of NaH2PO4 is
A. 25.0 cm3
B. 50.0 cm3
C. 75.0 cm3
D. 100.0 cm3
Question 15
1 dm3 of a solution contains 13.25g of a mixture of anhydrous sodium carbonate
and sodium chloride. 25 cm3 of the solution requires 25 cm3 of 0.1M hydrochloric
acid for complete reaction. The percentage by mass of sodium carbonate in the
mixture is
A. 20%
B. 40%
C. 60%
D. 80%
Chem MC
2010-10-16 10:04 pm
回答 (1)
2010-10-17 7:20 am
✔ 最佳答案
Question 10The answer is: A. 0.120M
Vol. of HNO3 of:
1st = 32.5 - 1.0 = 31.5 cm³ (rejected)
2nd = 32.5 - 2.6 = 29.9 cm³
3rd = 33.2 - 3.1 = 33.1 cm³
Average vol. of HNO3 = (29.9 + 30.1)/2 = 30.0 cm³
No. of moles of HNO3 = 0.10 x (30.0/1000) = 0.003 mol
No. of moles of NaOH = 0.003 mol
Molarity of NaOH = 0.003 / (25/1000) = 0.120 M
=====
Question 11
The answer is: B. 0.800M
Vol. of HCl of:
1st = 21.60 - 1.00 = 21.50 cm³ (rejected)
2nd = 21.60 - 1.50 = 20.10 cm³
3rd = 21.80 - 1.90 = 19.90 cm³
Average vol. of HCl = (20.10 + 19.90)/2 = 20.00 cm³
No. of moles of HCl = 1.00 x (20.00/1000) = 0.02 mol
No. of moles of NaOH = 0.02 mol
Molarity of NaOH = 0.02 / (25/1000) = 0.800 M
=====
Question 12
The answer is: D. 10
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
No. of moles of HCl = 0.1 x (12.5/1000) = 0.00125 mol
No. of moles of Na2CO3 = 0.00125/2 = 0.000625 mol
No. of moles of 1.43 g of Na2CO3•nH2O = 0.000625 x (200/25) = 0.005 mol
Molar mass of Na2CO3•nH2O = 1.43/0.005 = 286 g mol¯¹
(23x2 + 12 + 16x3) + n(1x2 + 16) = 286
n = 10
=====
Question 13
The answer is: A. 40.0 cm3 of H3PO4 (a tribasic acid)
No. of moles of NaOH needed:
A. 0.5 x (40/1000) x 3 = 0.06 mol
B. 0.5 x (45/1000) x 2 = 0.045 mol
C. 0.5 x (45/1000) x 2 = 0.045 mol
D. 0.5 x (90/1000) x 1 = 0.045 mol
The greatest amount of NaOH is needed in A.
=====
Question 14
The answer is: B. 50.0 cm³
NaOH + H3PO4 → NaH2PO4 + H2O
No. of moles of H3PO4 = 0.2 x (25/1000) = 0.005 mol
No. of moles of NaOH = 0.005 mol
Vol. of Na2CO3 solution = 0.005/0.1 = 0.050 dm³ = 50.0 cm³
=====
Question 15
The answer is: B. 40%
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
No. of moles of HCl used in titration = 0.1 x (25/1000) = 0.0025 mol
No. of moles of Na2CO3 used in titration = 0.0025 x (1/2) = 0.00125 mol
No. of moles of Na2CO3 in the mixture = 0.00125 x (1000/25) = 0.05 mol
Mass of Na2CO3 in the mixture = 0.05 x (23x1 + 12 + 16x3) = 5.3 g
% by mass of Na2CO3 in the mixture = (5.3/13.25) x 100% = 40%
參考: 土扁
收錄日期: 2021-04-13 17:34:48
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