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calculation in analytical chemistry
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不好意思,这些多我来说完全是新东西
1.what is the formal concentration express in mol/L of NaCl when 32.0g are dissolve in water and diluted to 500 ml
2.The concentration of alkane C20H42 (FW=282.55) in a particular sample of rain water is 0.2ppb . if the density of rain water is closed to 1.00gm/L . Find the molar concentration of the alkane.
3.how many of perochloric acid (HClO) are contain in 37.6g of 70.5% wt aqueous perochloric acid ? how many grams of water are in the same solution.
4.Consider a reservoir with a diameter of 450m and a depth of 10.0m . How many grams of F- should be added to give 1.6ppm ? The Flouride is provided by hydrogen hexaluorosilicate,H2SIF6.how many grams of hexaluorosilicate contain in this much F-
5.how many grams of 50% wt NaOH should be diluted to 1.0 L to make 0.1 M NaOH ?
6.How many ml of concentrated sulphuric acid labeled 98 wt % to prepare 1 L of 1 M H2SO4 ?
7. What is the density of 53.4 % wt aq NaOH if 16.7ml of the solution diluted to 2.0 L to give 0.169M NaOH ?
我试过很多次了,但是还是找不到答案 ,lecturer assume 我们会,所以没教我们 ... 希望大家可以帮忙我 |
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发表于 25-3-2012 02:34 PM
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本帖最后由 nkrealman 于 26-3-2012 04:59 PM 编辑
1. 1 gmol NaCl 有 58.4430g...32g 有 0.5475....dissolve in 0.5 L = 1.10 g/ L =1.10 M
2. 1L of rain water = 1000g ,the mass concentration of alkane is 0.2 X 10^-9 X 10^3
找到的答案住除FW 就找到 molar concentration 了 7.07 X10^-10 M
3.1 g of mixture containg 0.705g of acid....so 37..6g mix contains =26.508g
same goes for water..1 g of mix contains 0.295g of water
請指正 |
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发表于 25-3-2012 04:10 PM
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这不是高一的化学?为什么你不会?你有去问人吗?
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楼主 |
发表于 25-3-2012 11:33 PM
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这不是高一的化学?为什么你不会?你有去问人吗?
颜清华 发表于 25-3-2012 04:10 PM
完全不会,完全不懂%wt , ppm , ppb 是什么 只是懂ppm=mg/L,然后不会做,中学只有 MV 和 conversion mol -> 之类而已 ...老师当我们会了 |
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楼主 |
发表于 26-3-2012 12:17 AM
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1. 1 gmol NaCl 有 58.4430g...32g 有 0.5475....dissolve in 0.5 L = 1.10 g/ L =1.10 M
2. 1L of rai ...
nkrealman 发表于 25-3-2012 02:34 PM
2. 我不明白什么是mass concentration , 怎样拿到0.32 ? |
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楼主 |
发表于 26-3-2012 01:02 AM
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不好意思,除了第5题其他的题目我也不会,ml 和 g 的算法不一样吗?第6题用ml |
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发表于 26-3-2012 03:12 PM
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将mass 和 volume convert 去 Molar
将 ppm, ppb 当成 % wt 看待
ppm = part per million, 1/1000000 -> %
ppb = part per billion, 1/1000000000 -> % |
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发表于 26-3-2012 05:13 PM
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回复 5# 白羊座aries
我不用 mass concentration 好了..ppb 是 part per billion,ppm 是 part per million..
也就是說,每 10^9 (g or mol ) of rain water containing 0.2 (g/mol) of alkane..
ppb 的 formula 是 = mass fraction/mol fraction x 10^9
也就是說 mass fraction of alkane is 0.2 x10^9
我們take basis =1 L of rain water
1 L of rain water =1000g rain water
從 ppb 我們可以找 mass of alkane containing in 1000g of rain water...2x 10^-7 g
g / MW = mol...再除 1 L of rain water 你就可以找到 多少 mol of alkane per L rain water 了 |
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发表于 26-3-2012 05:16 PM
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本帖最后由 nkrealman 于 26-3-2012 05:43 PM 编辑
aqueous perochloric acid ,我們 assume only contain acid and water...
70.5% wt of acid....就是說 acid 的重量佔了 70.5 %
所以,在 37.6 g of aqueous perochloric acid,有 70.5 % 是屬於 acid 的,剩下的是 water
第 4 題我也不是很懂。。
第5題, 1L 0.1 M of aqueous NaOH 有 0.1 mol of NaOH....也就是說有 4g...因為 wt 只是 50%....所以要用 8g of solution 才有 4g of NaOH
第六題是不是少給 information?應該還欠 M of the concentrated acid..或者是 density
第 7 題M1V1 = M2V2
M1 = molarity of conc. NaOH (unknown)
V1 = volume of conc. NaOH
M2 = molarity of diluted NaOH
V2 = volume of diluted NaOH
M1 = (M2V2 / V1) = (0.169)(2000 mL) / (16.7 mL) = 20.2 M
20.2 moles NaOH / L x (40.0 g NaOH / 1 mole NaOH) = 808 g NaOH / L
53.4 mass % NaOH = 53.4 g NaOH / 100 g solution
How much solution contains 808 g NaOH?
(53.4 g NaOH / 100 g soln.) = (808 g NaOH / x g solution). By proportion, x = 1510 g
So 1510 g = 1L, D = g/mL = 1510 g / 1000 mL = 1.510 g/mL |
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发表于 30-3-2012 02:00 PM
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本帖最后由 cl88yee 于 31-3-2012 12:11 AM 编辑
不好意思,这些多我来说完全是新东西
1.what is the formal concentration express in mol/L of NaCl when ...
白羊座aries 发表于 25-3-2012 11:37 AM
第四题是什么答案? 我算到 grm of F- = 2544.690 grm..... grm of H2SIF6= 6138.5grm......对吗? |
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发表于 30-3-2012 02:34 PM
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这些问题是University level or secondary skul level?
还蛮难的。。。。 |
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发表于 30-3-2012 03:12 PM
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aqueous perochloric acid ,我們 assume only contain acid and water...
70.5% wt of acid....就是說 ac ...
nkrealman 发表于 26-3-2012 05:16 PM
第六题应该能做。。。。
1mol in 1L need 1 mol
mol = mass/molar mass
molar mass of H2SO4 = (2+32+(16X4))
1 mol = 0.98Y/ 96.04
Y= 98 grm
Assume that 1 grm = 1 ml, then 98 grm = 98ml.....
有错请指正。。。。 |
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楼主 |
发表于 1-4-2012 03:42 AM
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第四题是什么答案? 我算到 grm of F- = 2544.690 grm..... grm of H2SIF6= 6138.5grm......对吗?
cl88yee 发表于 30-3-2012 02:00 PM
答案是 2.5x106
; 3.2x106
是大学first year 的chem 来的 |
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发表于 1-4-2012 10:32 AM
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答案是 2.5x106
; 3.2x106
是大学first year 的chem 来的
白羊座aries 发表于 1-4-2012 03:42 AM
介意把jalan kerja放上来吗? |
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楼主 |
发表于 2-5-2012 02:08 PM
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A containing 500ml of 0.0035 mole of Fe3+ is precipitated to Fe(OH)3 with Ksp=1.1x10^-36
a)calculate its solubility
b)calculate percentage of Fe3+ will be precipitate if the pH is adjusted to 2.
a的是不是 1.1x10^-36=( s )( 3s )^3
然后算s?
b的我完全没有idea |
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