JEE Challenger
More from Solutions
ChemMedAdv 2012 • Paper 1

Volume of HCl Stock Solution Required for Preparation

A stock solution of HCl\text{HCl} with 29.2% (w/w)29.2\%\text{ (w/w)} concentration has a density equal to 1.25 g mL−11.25\text{ g mL}^{-1}. Given that the molecular weight of HCl\text{HCl} is 36.5 g mol−136.5\text{ g mol}^{-1}, calculate the volume (in mL\text{mL}) of stock solution needed to prepare a 200 mL200\text{ mL} solution of 0.4 M HCl0.4\text{ M HCl}.

Official Numerical Answer8

Topics & Concepts

SolutionsConcentration Terms

Step-by-Step Solution

To find the required volume of the HCl\text{HCl} stock solution, we first calculate its molarity and then use the dilution law.

Step 1: Calculate the molarity of the stock solution (M1M_1)

Let us consider 100 g100\text{ g} of the stock solution:

  • Mass of HCl\text{HCl} solute =29.2 g= 29.2\text{ g}
  • Density of the solution (dd) =1.25 g mL−1= 1.25\text{ g mL}^{-1}
  • Molar mass of HCl\text{HCl} (MHClM_{\text{HCl}}) =36.5 g mol−1= 36.5\text{ g mol}^{-1}

The volume of 100 g100\text{ g} of solution is: Volume of solution=MassDensity=100 g1.25 g mL−1=80 mL=0.08 L\text{Volume of solution} = \frac{\text{Mass}}{\text{Density}} = \frac{100\text{ g}}{1.25\text{ g mL}^{-1}} = 80\text{ mL} = 0.08\text{ L}

The number of moles of HCl\text{HCl} present in 100 g100\text{ g} of solution is: Moles of HCl=29.2 g36.5 g mol−1=0.8 mol\text{Moles of HCl} = \frac{29.2\text{ g}}{36.5\text{ g mol}^{-1}} = 0.8\text{ mol}

Therefore, the molarity of the stock solution (M1M_1) is: M1=Moles of soluteVolume of solution in L=0.8 mol0.08 L=10 MM_1 = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{0.8\text{ mol}}{0.08\text{ L}} = 10\text{ M}

(Alternatively, using the formula M=%(w/w)×d×10MHCl=29.2×1.25×1036.5=10 MM = \frac{\%(\text{w/w}) \times d \times 10}{M_{\text{HCl}}} = \frac{29.2 \times 1.25 \times 10}{36.5} = 10\text{ M})

Step 2: Calculate the required volume of the stock solution (V1V_1)

We need to prepare V2=200 mLV_2 = 200\text{ mL} of a M2=0.4 MM_2 = 0.4\text{ M} solution.

Using the dilution equation: M1V1=M2V2M_1 V_1 = M_2 V_2

Substituting the known values: 10 M×V1=0.4 M×200 mL10\text{ M} \times V_1 = 0.4\text{ M} \times 200\text{ mL}

V1=0.4×20010=8 mLV_1 = \frac{0.4 \times 200}{10} = 8\text{ mL}

Thus, the volume of stock solution needed is 8 mL8\text{ mL}.

PreviousNext