Home
Class 12
CHEMISTRY
29.2% (w/w) HCl stock solution has a den...

29.2% (w/w) HCl stock solution has a density of 1.25 g `mL^(-1)` . The molecular weight of HCl is 36.5 g `"mol"^(-1)`.
Find the Volume (V)(mL) of stock solution required to prepare a 500 mL solution of 0.4 M HCl.
Report your answer as `V//5`

Text Solution

Verified by Experts

The correct Answer is:
4
29.2% (w/w) HCl has density=1.25g/ml
Now, mole of HCl required in 0.4 M HCl
`=0.4xx0.5` mole =0.2 mole
If v mol of orginal HCl solution is taken
then volume of solution =1.25 v
mass of HCl=(1.25 v x 0.292)
mole of HCl `=(1.25vxx0.292)/36.5=0.2`
so, `V=(36.5xx0.2)/(0.29xx1.25)` mol =20 mL
V/5=4
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • QUALITATIVE ANALYSIS PART 1

    RESONANCE|Exercise A.L.P|39 Videos

  • SOLID STATE

    RESONANCE|Exercise PHYSICAL CHEMITRY (SOLID STATE)|45 Videos

Similar Questions

Explore conceptually related problems

29.2% (w/w) HCl stock solution has a density of 1.25 g mL^(-1) . The molecular weight of HCl is 36.5 g "mol"^(-1) . The volume (mL) of stock solution required to prepare a 200mL solution of 0.4 M HCl is :

29.2% (w//w) HCl stock, solution has a density of 1.25 g mL^(-1) . The molecular weight of HCl is 36.5 g mol^(-1) . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is :

Knowledge Check

  • 29.2% (W//W) HCl stock solution has density of 1.25g mL^(-1) . The molar mass of HCl is 36.5g mol^(-1) . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4M HCl is

    A
    `8 mL`
    B
    `6 mL`
    C
    `7 mL`
    D
    `5 mL`

  • An aqueous solution has a density of 1. 37 g//ml . If molecular wt of solute is 68.5 g//mol , then calculate molarity of solution.

    A
    `2 M`
    B
    `50 M`
    C
    `5 M`
    D
    `20 M`

  • The density of 2.0 M solution of a solute is 1.2 g mL^(-1) . If the molecular mass of the solute is 100 g mol^(-1) , then the molality of the solution is

    A
    2.0 m
    B
    1.2 m
    C
    1.0 m
    D
    0.6 m

    • Similar Questions

    Explore conceptually related problems

    29.9% (w/w) HCl stock solution has a density of "1.25 g mL"^(-1) . The molecular weight of HCl is "6.5 g mol"^(-1) . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is

    29.2 % (W//W) HCl stock solution has a density of 1.25 "g mL"^(-1) . The molecular mass of HCl is 36.5 g mol^(1) . What is the volume (mL) of stock required to prepare 200 mL of solution of 0.4 M HCl ?

    29.2% (W/W) HCI stock solution has a density of 1.25 g mL^(-1) . The molecular mass of HCI is 36.5 g mol^(-1) Calculate the volume (mL) of stock solution required to prepare 200 mL of 0.4 M HCI

    The answer to each of the following questions is a single digit integer, ranging from 0 to 9. If correct answers to the question number A,B,C and D (say) are 4,0,9 and 2 respectively, then correct darkening of bubbles should be as shown on the side. (D) 29.2 (w/w) HCl stock solution has a density of 1.25 g mL^(-1) . The molecular weight of HCl is 36.5 g mol^(-1) . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is ..........................

    The density of 2.0 M solution of a solute is 1.2 g mL^(-1) . If the molecular mass of the solute is 100 g mol^(-1) , then the molality of the solution is :

    Home
    Profile