An old sodium thiosulphate (na2s2o3, 158.2 g/mol) solution is labelled 0.241 m. to verify this, during a redox titration of 0.141 l of this solution, 43.22 ml of 0.264 m iodine solution were needed to reach the equivalence point. what is the percentage of sodium thiosulphate found with respect to that labelled?

First up write the aim
then write the materials reqd.
then the procedure which you follow
then observation
then conclusion
then you can write lab safety rules or precautions(you can avoid if u want).
lemme' give an example
aim: to observe the double displacement reaction
materials reqd: bacl2 and na2so4, test tubes and a dropper
procedure:
1. take bacl2 in a test tube and na2so4 in the other.
2. then pour bacl2 into na2so4 solution.
3. observe
observation:
you will see that a white precipitate is formed called baso4. along with that common salt or nacl is formed.
the equation is
bacl2+na2so4=baso4+2nacl
conclusion:
a double displacement reaction is observed.
hope it

It would be c. risk.

Moles of co₂ = 2.35/44 = 0.0534 mole.from ideal gas equation pv=nrt,  t = 18+273k = 291 kelvin.∴px5 = 0.0534x0.0821x291∴p = 0.0534x0.0821x291/5 = 0.2551 atm.now, 1 atm = 760 mm of hg  ∴ 0.02551 atm = 0.02551 x 760 = 19.38 mm of hg