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Concertrated hydrochloric acid is 36% ( W/V) HCL and has a density of 1.18gcm-3

If 10cm3 of the concertrated acid is dissolved in water to produce 1dm3 of solution, Calculate the concertration in moldm3 of the solution
If 5cm3 of the dilute above is made up 20cm3 what will be the new solution in moldm-3

1 answer

To calculate the concentration of the diluted hydrochloric acid solution in mol/dm³ (molarity), we can follow these steps:

Step 1: Calculate the mass of HCl in the concentrated solution.

  1. Volume of concentrated HCl solution = 10 cm³
  2. Density of concentrated HCl = 1.18 g/cm³
  3. Percentage concentration (w/v) = 36%

Using the formula for mass based on density:

\[ \text{mass of concentrated solution} = \text{volume} \times \text{density} = 10 , \text{cm}^3 \times 1.18 , \text{g/cm}^3 = 11.8 , \text{g} \]

Next, we can find out how much of this mass is HCl:

\[ \text{mass of HCl} = 11.8 , \text{g} \times \frac{36}{100} = 4.248 , \text{g} \]

Step 2: Convert the mass of HCl to moles.

  1. Molar mass of HCl = 1.01 + 35.45 = 36.46 g/mol

\[ \text{moles of HCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.248 , \text{g}}{36.46 , \text{g/mol}} \approx 0.1165 , \text{mol} \]

Step 3: Calculate the concentration of the solution after dilution.

The total volume of the solution after dilution is 1 dm³ (which is equal to 1000 cm³).

\[ \text{Concentration (C)} = \frac{\text{moles of solute}}{\text{volume of solution in dm}^3} = \frac{0.1165 , \text{mol}}{1 , \text{dm}^3} = 0.1165 , \text{mol/dm}^3 \]

Step 4: Calculate the concentration of the new diluted solution.

Now, if 5 cm³ of this dilute solution is further diluted to a total volume of 20 cm³:

  1. Initial concentration of dilute solution = 0.1165 mol/dm³
  2. Volume taken = 5 cm³ = 0.005 dm³
  3. Final volume after dilution = 20 cm³ = 0.020 dm³

Using the dilution formula:

\[ C_1V_1 = C_2V_2 \]

Where:

  • \(C_1\) = initial concentration (0.1165 mol/dm³)
  • \(V_1\) = initial volume (0.005 dm³)
  • \(C_2\) = final concentration (unknown)
  • \(V_2\) = final volume (0.020 dm³)

Rearranging gives:

\[ C_2 = \frac{C_1V_1}{V_2} = \frac{0.1165 , \text{mol/dm}^3 \times 0.005 , \text{dm}^3}{0.020 , \text{dm}^3} \]

Calculating \(C_2\):

\[ C_2 = \frac{0.0005825 , \text{mol}}{0.020 , \text{dm}^3} = 0.029625 , \text{mol/dm}^3 \approx 0.0296 , \text{mol/dm}^3 \]

Final Answers:

  1. Concentration of the first diluted solution: 0.1165 mol/dm³
  2. Concentration of the new solution after the second dilution: 0.0296 mol/dm³