Home
Class 12
CHEMISTRY
29.2% (W//W) HCl stock solution has dens...

`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

Text Solution

Verified by Experts

The correct Answer is:
`(8mL)`

Mass of `HCl` in `1.0 mL` stock solution
`= 1.25 xx (29.2)/(100) = 0.365 g`
Mass of `HCl` required for `200 mL 0.4 M HCl`
`= (200)/(1000) xx 0.4 xx 36.5 = 0.08 xx 36.5 g`
`:. 0.365 g` of `HCl` is present in `1.0 mL` stock solution.
`0.08 xx 36.5 g HCl` will be present in `(0.08 xx 36.5)/(0.365) = 8.0 mL`
Promotional Banner

Topper's Solved these Questions

  • SOME BASIC CONCEPTS OF CHEMISTRY

    IIT-JEE PREVIOUS YEAR (CHEMISTRY)|Exercise Subjective Questions|2 Videos
  • SOLUTIONS AND COLLIGATIVE PROPERTIES

    IIT-JEE PREVIOUS YEAR (CHEMISTRY)|Exercise Assertion-Reason Type|2 Videos
  • STATES OF MATTER

    IIT-JEE PREVIOUS YEAR (CHEMISTRY)|Exercise INTEGER_TYPE|2 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 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 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 :

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 ..........................

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

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

A 30% (W/W) aqueous solution has density 2gm/ml and molarity 2M, then molar mass of solute is:

Calculate the volume of 0.015 M HCl solution required to prepare 250 mL of a 5.25 xx 10^(-3) M HCl solution

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