If 11.5 ml of vinegar sample (d=1g/ml) is titrated with 18.5 ml of standardized Sodium hydroxide
solution. What is the percentage of acetic acid (by mass) in the vinegar

4.83% of acetic acid in the vinegar

Explanation:

The neutralization reaction of acetic acid (CH₃COOH) with sodium hydroxide (NaOH) is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

Assuming the sodium hydroxide solution has a concentration of 0.500M, moles used in the neutralization reaction are:

0.0185L × (0.500mol / L) = 0.00925 moles of NaOH = 0.00925 moles CH₃COOH

Because 1 mole of acid reacts per mole of NaOH

Using molar mass of acetic acid (60g/mol):

0.00925moles CH₃COOH ₓ (60g / mol) = 0.555g of CH₃COOH

As mass of vinegar sample is 11.5g (d = 1g/mL), percentage of acetic acid by mass is:

0.555g CH₃COOH / 11.5g vinegar × 100 = 4.83% of acetic acid in the vinegar

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.