You titrated a 25.00 ml solution of 0.02 m oxalic acid with a freshly prepared solution of kmno4. if it took 42.32 ml of this solution to reach the endpoint, what was the molarity of the kmno4?

This is a Redox Titration reaction and the overall ionic equation is given as follows;

2 MnO⁴⁻ + 5C₂O₄ ²⁻+ 16H+  →  2Mn²⁺ + 10CO₂ + 8H₂O

From here, we could see that the stoichiometry ratio of KMnO₄ and oxalic acid = 2:5

To calculate the molarity  of KMnO₄, we will use the following equation.

x₁M₁V₁= x₂M₂V₂                                      (1)

Where,

x₁ = 2, stoichiometry number of KMnO₄ in the balanced reaction

x₂ = 5,  stoichiometry number of C₂O₄ in the balanced reaction

M₁ and M₂  are the molarities of oxalic acid and KMnO₄ solutions

V1  and V2  are the volumes of oxalic acid and KMnO₄ solutions.

Making M₂, the subject of the formula and substituting the given values, we have,

2M₁V₁ = 5M₂V₂, M₂=2M₁V₁ / 5V₂

= = 2x0.02Mx25mL/ 5x42.32mL

M₂=0.0047M

Hence, the molarity of the KMnO4 at endpoint =0.0047M

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.