how many millimeter should be taken out to prepare 0.5 normal HCL solution of 200 ml. density=1.18, purity 37%

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One mole of HCl is required to prepare 1 L of 1N solution.

Calculate number of moles (n) required to prepare 200/1000 L of 0.5N solution.

Find molecular mass M of HCl in g

mass of pure HCl required = nM g

Volume of given acid required
= nM ÷ (1.18*0.37) g÷(g/mL)

Here is slightly different format.

First determine the N of the HCl. The molar mass HCl is about 36.5 g/mol.
1.18g/mL x 1000 mL x 0.37 x 1/36.5 = ? and that is approx 12N
Then use the dilution formula of
mL1 x N1 = mL2 x N2
mL1 x 12 = 200 x 0.5
Solve for mL1.

To prepare a 0.5 Normal HCl solution of 200 ml, you need to know the molar concentration of the hydrochloric acid (HCl) solution and the desired final volume.

First, let's calculate the number of moles (n) of HCl required using the formula:

n = C * V

where C is the molar concentration and V is the volume in liters.

Since the density of the given HCl solution is 1.18 g/ml and the purity is 37%, we need to calculate the actual concentration of HCl in the solution:

Concentration of HCl = Purity * Density * (1 / Molar Mass of HCl)

The molar mass of HCl is approximately 36.46 grams per mole.

Concentration of HCl = 0.37 * 1.18 g/ml * (1 mol HCl / 36.46 g)

Now, let's calculate the number of moles of HCl required:

n = 0.37 * 1.18 * (1 / 36.46) * 200 / 1000

The factor of 200/1000 is used to convert the desired volume from milliliters to liters.

Finally, we can calculate how many milliliters of the given HCl solution needs to be taken out. To find this, we divide the number of moles required (calculated above) by the molar concentration of the given HCl solution:

Volume to be taken out (in mL) = (n / C) * 1000

Now you can substitute the values and calculate the answer:

Volume to be taken out = (n / C) * 1000

Make sure to replace 'C' with the actual molar concentration calculated above and 'n' with the number of moles required.

Please note that the actual concentration of the given HCl solution was not provided in the question, so it may not be possible to calculate the final answer with the given information.