How do I prepare 250 ml of 0.2 N H2SO4 from concentrated H2 SO4 solution with 98% by weight d = 1.84 g/ml?

Molar mass of H2SO4 = 98 g/mol.

To prepare 1000 mL 0.1M H2SO4 solution 0.1 mole H2SO4 = (98×0.1) g = 9.8 g H2SO4 is required.

So, to prepare 250 mL 0.1N H2SO4, (250/1000)×9.8 = 2.45 g H2SO4 is required.

Procedure: Take a 250 mL volumetric flask. Pour about 150 mL distilled water in it. Then pour 2.45 g of H2SO4 in it with constant stirring. Make up the volume with distilled water up to the 250 mL mark. Your desired solution is ready.

Hope, this helps.

Here,normality,N= 0.02

Molar mass of H2SO4,M= 98g

Basicity of H2SO4,n = 2

Volume of the solution,V= 500ml= 0.5L

Mass of H2SO4,W=?

We know,

S= nw/MV

or,W= MNV/n

= 98×0.02×0.5/2

= 0.49g

i.e.0.49g H2SO4 will be taken in a 500 ml volumetric flask.

I think your question should be: How do I prepare 0.5 M solution from H2SO4 having a density of 1.84 g/mL and 98 % purity?

The concentrated acid would have a molarity of 18.4 M; viz:

1.84 g/mL x 0.98 x 1000 mL/L x 1 mol/98 g = 18.4 mol/L or 18.4 M

You did not state the volume of dilute solution required. So, let us say you want 500 mL of 0.5 M.

Using the “dilution equation”, Cc x Vc = Cd x Vd

18.4 M x Vc = 0.5 M x 500 mL

Vc = 13.6 mL

So, 13.6 mL of 18.4 M solution is diluted to 500 mL total volume to give a 0.5 M solution (as an example).

N1V1 = N2 V2

0.250 x 0.2 = 36.8V2

V2 = 1.36 mL

Take this volume and dilute til with wáter till a volume o 250 mL carefully

(1.84 g solution/ 1 mL solution) x (98 g H2SO4/ 100 g solution) x (1 mole H2SO4/ 98.09 g H2SO4) x 1000 mL solution = 18.4 mole H2SO4 /L solution

(18.4 mole H2SO4 /L solution) x (2 equivalents H+/ 1 mole H2SO4) = 36.8 Equivalents/L solution

V(1) x N(1) = V(2) x N(2)

V(2) = [250 mL x (0.2 equivalents/Liter)]/(36.8 equivalents/Liter) = 1.36 mL

So, measure 1.36 mL and place into a 250 mL volumetric flask and add distilled water to the mark on the neck of the flask.

When we dilute a concentrated acid, there is an over-riding issue of safety…

$ext{If you spit in acid, it SPITZ back!!!}$

And so when conc. acid, sulfuric or hydrochloric etc, is DILUTED, it is ALWAYS acid to water … and NEVER the reverse…

And so let us see if we can address the question on these terms…

We required $0.2•N$ $H_{2}SO_{4}$ … and this is more or less $0.1•mol•L^{-1}$ $H_{2}SO_{4}$ because sulfuric acid is a DIACID...

TO begin $[H_{2}SO_{4}]=dfrac{frac{ ext{98%}×1.84•g:^{-1}}{98.08•g•mol^{-1}}}{0.001•L}=18.38•mol•L^{-1}$

And we require $250•mL×0.1•mol•L^{-1}=0.0250•mol$

And so we needs $dfrac{0.0250•mol}{18.38•mol•L^{-1}}×1000•mL•L^{-1}=1.36•mL$

…and REMEMBER acid to water…

$ext{May her rest be long and placid}$

$ext{She added water to the acid}$

$ext{The other girl did as she oughter…}$

$ext{She added acid to the water…}$

we know,

normality(N) = molarity(M) * no. of transferable ion

M = N / no. of transferable ion

M= 0.5 / 2 = 0.25

again we know , V1M1 = V2M2

given, V1 = 1000 ml

M1 = 0.25 mole per liter

M2 = Most of the stock solutions of H2SO4 solution having 18M concentration

V2 = ?

V2 = V1M1 /M2 = (0.25 * 1000) / 18 = 13.88 ml

ANSWER: 13.88 ml

0.36 ml sulfuric acid in 100 ml water

I understand that you want to prepare 250 mL of 1 M H2SO4. You have available an adequate volume of 5 M H2SO4.

Use the formula

M1 x V1 = M2 x V2

1 M x 250 mL = 5 M x V2

V2 = 1 x 250 / 5 = 50 mL

Carefully mix 200 mL of water and 50 mL of 5 M H2SO4.

To check: Ask how many moles of acid are required to prepare the 1 M solution? Answer 0.25 moles.

And how many moles did we obtain from the 5 M solution? Answer 0.25 moles

QED

0.050 L of 5 M acid → 0.25 moles

0.250 L of 1 M acid → 0.25 moles

Dear anonymous,

Apparently, you want us to do your homework for you. Below is the solution to your question. Do know that this solution is useless to you unless you understand each step and why it is taken. So while I give you the solution to the question, it will be up to you to study this answer carefully and make sure you understand it.

To start of with, we will need to fix the question itself as it contains a rather big error. The density of any substance is given in g/cm^3. Not in g/mol as that would be the molecular weight. So the density is 1.84 g/cm^3 (=1.84 g/mL).

The answer:

0.2M = 0.2 mole/L

Therefore, 0.5L * 0.2M = 0.1 mole of H2SO4 required.

The molar mass of H2SO4 is 98.079 g/mole (see periodic table or google).

Therefore, 98.079 g/mole * 0.1 mole = 9.8079 g

9.8079 g/ 1.84 g/cm^3 = 5.33038 cm^3 (=5.33038 mL)

The purity of your substance is 96%, therefore you will actually need:

5.33038*(1/0.96) = 5.55248 mL.

The required precision in your answer is 3 digits (due to the precision of 500 mL).

Therefore, the answer is 5.55 mL of acid and (500–5.55 =) 494 mL of water.

Note that when preparing such a dilution, always add acid to water. Never the other way around. Pure H2SO4 is an extremely dangerous substance. Handle only under the supervision of a professional.