commercial vinegar contains ethanoic acid.50cm3 of vinegar are diluted and made up to250cm3. 25cm3 portions of diluted vinegar are titrated with 0.2M sodium hydroxide solution,the volume of NaOHis 20cm3
1.calculate the concentration of ethanoic acid in the vinegar in mol dm-3
2.calculate the concentration of ethanioc acid in g dm-3
3.assume that the density of vinegar is 1g m-3,calculate the percentage by mass of ethanioc acid in the vinegar
vita-c is a tablet used to supplement vitamin C to people's diet.the tablet contains a monobasic acid called ascorbic acid.the molar mass of the acid is 176g mol-1.a student disslove vita-C(mass-6.00g)in water to form a solution. the solutin requires 15.0cm3 of 1M sodium hydroxide solution for complete reaction,calculate the mass of the acid in vita-C
chemical calculation
2008-05-03 12:47 am
回答 (2)
2008-05-03 2:16 am
✔ 最佳答案
(commercial vinegar)1.
Consider the titration:
CH3COOH + NaOH → CH3COONa + H2O
No. of moles of NaOH used = MV = 0.2 x (20/1000) = 0.004 mol
No. of moles of CH3COOH = 0.004 mol
Volume of diluted vinegar used = 25 cm3 = 0.025 dm3
Concentration CH3COOH in the diluted vinegar
= mol/V
= 0.004/0.025
= 0.16 mol dm-3
Consider the dilution of vinegar:
Original vinegar: M1 = ?, V1 = 50 cm3
Diluted vinegar: M2 = 0.16 mol dm-3, V2 = 250 cm3
M1V1 = M2V2
M1 x 50 = 0.16 x 250
Concentration of CH3COOH in the vinegar, M1 = 0.8 mol dm-3
2.
Molar mass of CH3COOH = 12x2 + 1x4 + 16x2 = 60 g mol-1
Concentration of CH3COOH in the vinegar
= (0.8 mol dm-3) x (60 g mol-1)
= 48 g dm-3
3.
Consider 1 dm3 (1000 cm3) of the vinegar.
Density of vinegar ρ = 1 g cm-3
(It should not be 1 g m-3.)
Total mass of the vinegar =ρV = 1 x 1000 = 1000 g
Mass of CH3COOH in the vinegar = 48 g
Percentage by mass of CH3COOH = (48/1000) x 100% = 4.8%
==========
(Vitamin C)
Denote ascorbic acid as HA (a monobasic acid).
HA + NaOH → NaA + H2O
Mole ratio HA : NaOH = 1 : 1
No. of moles of NaOH used = MV = 1 x (15/1000) = 0.015 mol
No. of moles of HA used = 0.015 mol
Mass of HA in vita-C = mass x (molar mass) = 0.015x 176 = 2.64 g
2008-05-03 2:23 am
NaOH + CH3COOH ------> CH3COONa + H2O
No. of mole of NaOH reacted = 0.2 x 20/1000 = 0.004
No. of mole of ethanoic acid reacted = 0.004
Concentration of ethanoic acid in vinegar
= (0.004 x 250/25) x 1000/50 = 0.8 M
Concentration of ethanoci acid
= 0.8 x (12+3+12+16x2+1)
= 48 g dm-3
3.assume that the density of vinegar is ,calculate the percentage by mass of ethanioc acid in the vinegar
Your question is not reasonable. I guess the deinsity of vinegar should be 1g cm-3.
48 g dm-3 = (48/1000)gdm-3 = 0.048 gdm-3
% by mass = 0.048/1 x 100% = 4.8
---------------------------------------------------------
Mole ratio of ascorbic acid to NaOH = 1:1
no. of mole of NaOH reacted = 1 x 15/1000 = 0.015
no. of mole of ascorbic acid reacted = 0.015
mass of ascorbic acid = 0.015 x 176g = 2.64g
No. of mole of NaOH reacted = 0.2 x 20/1000 = 0.004
No. of mole of ethanoic acid reacted = 0.004
Concentration of ethanoic acid in vinegar
= (0.004 x 250/25) x 1000/50 = 0.8 M
Concentration of ethanoci acid
= 0.8 x (12+3+12+16x2+1)
= 48 g dm-3
3.assume that the density of vinegar is ,calculate the percentage by mass of ethanioc acid in the vinegar
Your question is not reasonable. I guess the deinsity of vinegar should be 1g cm-3.
48 g dm-3 = (48/1000)gdm-3 = 0.048 gdm-3
% by mass = 0.048/1 x 100% = 4.8
---------------------------------------------------------
Mole ratio of ascorbic acid to NaOH = 1:1
no. of mole of NaOH reacted = 1 x 15/1000 = 0.015
no. of mole of ascorbic acid reacted = 0.015
mass of ascorbic acid = 0.015 x 176g = 2.64g
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