Concentrated H2SO4 is 96 %.( In laboratory )
As density of concentrated H2SO4 is 1.84gm/ml we will need this number as well, and as the Atomic Mass of H2SO4 is 98.08,as it is dibasic for normality it is 49
hence,
Calculation=((96/100)(1000)(1.84))/49=36.04
If H2SO4 concentrated is 36.04 M then for make a 1L solution of 1M H2SO4
(36.04)X (x) = 1X(1)
x = 1 X(1) / (36.04)
x=0.0277gm/ml of water
x = 27.7 mL of 36M H2SO4 per liter
Hence for 1N H2SO4 dissolve 27.7ml of it to 1000ml of solvent(Water)
that means for 0.1 N H2SO4 2.77 ml of it to 1000mL of solvent.
- transfer approx. 75 mL of distilled (deionized) water in a clean and dried 100 mL volumetric flask
- weight the flask
- add 0,4902 g of concentrated sulfuric acid
- stirr vigourosly for mixing
- put the flask in a thermostat at 20 0C
- after the flask with the solution reach 20 0C add easily distilled (deionized) water to the mark
- again stirr vigourosly
- transfer the solution from the volumetric flask in another glass bottle
- label the bottle (type of solution, concentration, date, operator)
Normality=Molarity x N-factor.
=>Molarity=Normality/N-factor.
N-factor for H2SO4 is 2.
So,
Molarity=0.02/2=0.01 M
This means we have to dissolve 0.01moles of H2SO4 in 1L of water t prepare a 0.02 N solution.
0.01 M H2SO4 = 98g/0.01=.98g
We dissolve .98g H2SO4 1L of water to get the desired solution
The preparation of 0.5 M sulfuric acid (H2SO4) depends on what you start with and what volume of 0.5 M solution you want. If you begin with concentration H2SO4, which is usually 36.8 N, then this is equal to 18.4 M. To prepare 1 liter of 0.5 M, you would proceed as follows:(x L)(18.4 M) = (1 L)(0.5 M) and x = 0.027 L = 27 mls.
So, take 27 mls of concentrated H2SO4 and dilute to a final volume of 1000 ml.
Assuming the conc. H2SO4 is 18 M (or close to it), then use V1M1 = V2M2 and to make, say 1 liter,(x ml)(18 M) = (1000 ml)(0.1 M) and x = 5.56 mls of conc. H2SO4 diluted to a final volume of 1 liter to make 0.1 M.
H2SO4 has a molar mass of 98.1 grams per mole. To prepare a 0.5 M solution, add 49.05 grams of sulfuric acid per liter of water.
take 14
ml of conc
sulfuric acid and add to volumetric flask contain 400 ml dist. water and complete to 1000 ml with water
suppose you have conc. sulfuric acid 96% sp.gr 1.84
so 1.25% sulfuric equal to 1/96*100=1.3
1.3ml of conc. sulfuric per 100ml water so 13 ml per liter
.98
The answer is 6,71 g dried KCl.
Remember M1V1=M2V2. (M is for molarity. V is for volume.)So:0.25M x V1 = 0.50 M x 0.050 LV1 = 0.50 M x 0.050L / 0.25 M = 0.1 LWait! You need it in mL so 0.1 L x 1000 = 100 mL
The molar weight of NaOH is 39.9971 g/mol. .15g is 3.75 millimoles. The reaction of H2SO4 and NaOH is H2SO4 + 2 NaOH --> Na2SO4 + 2 H2O. So the molar ratio of H2SO4 to NaOH is 1:2. So it takes 1.825 millimoles of H2SO4. Volume = molars/concentration. Then it takes 14.96 mL of H2SO4.
The molarity of a solution made by dissolving 23,4 g of sodium sulphate in enough water to make up a 125 mL solution is 1,318.
0.0125
125 ml 500(ml) * 0.05 = 25 25 / 0.20 = 125
No such thing as HSO in chemistry. If you're referring to H2SO4, which is sulfuric acid, then 125 grams of it would be: H2SO4 = 98g/mol; 98/1=125/x; solve for x to get about 1.28 moles.
125 percent into a percentage = 125%
The answer is 6,71 g dried KCl.
98 is 78.4 percent of 125 122.5 is 98 percent of 125
125 percent percent of 48 = 0.6
125 percent of 215000 = 268750
Answer this question…What is 75 of 125 as a percent?
125% = 1.25
92% = 0.920.92 * 125 = 115
The percentage of a 1 to 125 solution is simply 1/125 or 0.008 or 0.8%.
30 percent of 125 = 125 *30/100 = 37.5 125 - 37.5 = 87.5