How much dil HCL (10% w/w) HCL specific gravity = 1.045 would be required to prepare 2000ml of a solution containing 0.005 molar HCL? HCL m.w.t = 36.5

Who try to solve for me. Please help

The right answer is 34.9ml

Please check your post for typos. The answer isn't 34.9 mL. Using your numbers.

10% ww HCl means 10g HCl/100 g solution.
mols HCl = g/molar mass = 10/36.5 = 0.274
Volume of 100 g soln = mass/density = 100/1.045 = 95.69 = 0.09569 L
M 10% soln = 0.274/0.09568 = 2.86M
M x L = M x L
0.005M x 2 L = 2.86 M x ?L
Solve for ?L. I obtained 0.0035L or 3.50 mL.

To solve this question, we need to use the formula:

number of moles (n) = concentration (C) x volume (V)

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

n = C x V
= 0.005 mol/L x 2 L
= 0.01 moles

Since we know the molar weight of HCl is 36.5 g/mol, we can convert the number of moles to grams of HCl:

mass (m) = n x molar weight (MW)
= 0.01 moles x 36.5 g/mol
= 0.365 g

Next, we need to find the mass percent (w/w) of the 10% HCl solution. A 10% w/w solution means that 10 grams of HCl is dissolved in 100 grams of the solution:

%w/w = (mass of solute / mass of solution) x 100
= (10 g / 100 g) x 100
= 10%

Now we can determine the mass of the HCl solution required:

mass of HCl solution = (mass of HCl / %w/w of HCl) x 100
= (0.365 g / 10%) x 100
= 3.65 g

Next, we need to convert the mass of the HCl solution to volume using specific gravity. Specific gravity is the ratio of the density of a substance to the density of water at a specific temperature. In this case, the specific gravity of the HCl solution is given as 1.045.

density (d) = specific gravity (SG) x density of water at 4°C (1 g/mL)
= 1.045 x 1 g/mL
= 1.045 g/mL

So, the volume of the HCl solution required is:

volume (V) = mass (m) / density (d)
= 3.65 g / 1.045 g/mL
= 3.496 mL

Rounding to the nearest decimal place, the volume of the 10% HCl solution required is 3.5 mL. However, the question specifies that we need to prepare 2000 mL of the solution. Therefore:

volume of 10% HCl solution required = (3.496 mL / 1000 mL) x 2000 mL
= 6.992 mL
≈ 6.99 mL

So, approximately 6.99 mL of the 10% HCl solution would be required to prepare 2000 mL of a solution containing 0.005 M HCl.