In order to reduce the percentage of a chemical in a solution, first you need to use the general dilution equation which is:

(C1)(V1)=(C2)(V2)

Whereby C1 and C2 are concentration of the chemical before and after the dilution (reduction of concentration), respectively. In your case, C1 is 37% and C2 is 5%. Assume that you have 100 ml of 37% HCl in water, and you want to dilute these 100 ml of HCl solution using water as diluent to achieve a solution at lower HCl concentration of 5%, the equation can be re-written as:

(37%)(100 ml) = (5%)(V2)

Solving for V2, the final volume (V2) = 740 ml

The final volume includes the initial volume, in other words : V2 = V1 + V(additional water)

In your case: 740 ml = 100 ml + V(additional diluent)

Thus, V(additional diluent) = 640 ml of water required

To put long story short, add 640 ml of water to 100 ml of a solution of 37% HCl in water and mix them well, to achieve a final solution with volume of 740 ml whereby the HCl is 5% only.

You do below calculation for example 100ml of 5% HCl

5%×100 ml=37% × V ml then V ml = 5×100/37= 13.53 ml

Take 13.5 ml 37% of HCl and make up volume 100ml

In order to reduce the percentage of a chemical in a solution, first you need to use the general dilution equation which is:

(C1)(V1)=(C2)(V2)

Whereby C1 and C2 are concentration of the chemical before and after the dilution (reduction of concentration), respectively. In your case, C1 is 37% and C2 is 5%. Assume that you have 100 ml of 37% HCl in water, and you want to dilute these 100 ml of HCl solution using water as diluent to achieve a solution at lower HCl concentration of 5%, the equation can be re-written as:

(37%)(100 ml) = (5%)(V2)

Solving for V2, the final volume (V2) = 740 ml

The final volume includes the initial volume, in other words : V2 = V1 + V(additional water)

In your case: 740 ml = 100 ml + V(additional diluent)

Thus, V(additional diluent) = 640 ml of water required

To put long story short, add 640 ml of water to 100 ml of a solution of 37% HCl in water and mix them well, to achieve a final solution with volume of 740 ml whereby the HCl is 5% only.

I would do similar to what Ajay explained.

so if you need 1% of 35% you need to:

divide (1X100)/ 35= 2.85 ml

This is the volume you have from 35%.

Then:P

100-2.85=97.15 mls would be the volume you need to add to your 35% to give you 100mls of 1%.

Is that clear?

Simin.

C1 x V1= C2 x V2

C1 is the initial concentration(37%) C2 is the diluted concentration(5%)

for example if you want to prepare 100ml of 5% solution you do the following:

37 x V1= 5 x 100 => V1=13.51ml

So you must take 13.5 ml out of your initial solution (37%) and add them to a 100 ml volumetric flask then add distilled water until the bottom of the meniscus is level with the horizontal line on the neck of the flask.

Unless stated otherwise, the concentration indicated in the label, should be assumed as given in wt%, i.e. (g HCl / g solution)·100%. The same applies to the indicated dilute concentration. The density of 37.0 wt% HCl (aq.) solution is 1.184 g/cm3 (20 ºC) (*) ― likely available from the label; that for 5.00 wt% HCl is 1.023 g/cm3 (20 ºC) (*). The intended dilute solution can be then volumetrically prepared, at approx. 20ºC, while considering the appropriate (volumetric) dilution factor, calculated from HCl mass balance to the dilution process:

0.370·1.184·Vi (g HCl /cm3) = 0.0500·1.023·Vf (g HCl /cm3)

(*) R.H. Perry, D.W. Green, J.O. Maloney (Eds.), "Perry's Chemical Engineers’ Handbook", 7th ed., McGraw-Hill, 1997; Table 2.57.

There is a general rule to dilute a solution (for both organic or inorganic solutions): C1*V1=C2*V2

However:

- C1 is the original concentration of the solution (or the material disolved in the solute);

-V1 is the volume of the solution that you need to take from your orginal solution;

-C2 is the desired concentration, i.e your target concentration; and

-V2 is the final volume that you want to have.

Supposing that C1, C2, and V2 are known, you can now calculate how much molume you need from the original solution to prepare your diluted one:

V1=(C2*V2)/C1

After calculating your V1, you take this volume (which is lower than V2) than you complete it with the solvant that you have until you reach V2.

There is another simple way similar to above calculations, by which you can calculate your required final concentration.

Lets suppose, you want to prepare 1000 ml of 5 % HCl solution from 37% concentrated HCl.

C1*V1=C2*V2

V1 = (5 * 1000)/(100) = 50 ml (Consider you have 100% HCl solution)

37 % HCl is present in = 50 ml

1 % HCl is present in = 50/37

100% HCl is present in = (50/37)*(100) = 135.13 ml

Take 135.13 ml of Concentrated 37 % HCl solution and make total volume upto 1000 ml to obtain final solution with 5% concentration.