Estimate the freezing point depression of 0.19 m C12H22O11(aq). The freezing point constant of water is 1.86 K · kg/mol.
Answer in units of ◦C.

0.3634 °C

Explanation:

In this question, we are tasked with calculating the freezing point depression of 0.19 M C12H12O11( sucrose)

Mathematically this can be calculated using the formula;

ΔTf = i * kf * m

where ΔTf is the freezing point depression

i is Van’t Hoff factor

kf is the molal freezing point depression for the solvent

m is the molality of the solution

From the question we identify the following;

Since sucrose is a non-electrolyte, i = 1 ( i refers to the number of moles of particles obtained when 1 mol of solute dissolves)

m is 0.19

kf = 1.86

substituting these values into the equation, we have

ΔTf = 1 * 1.86 * 0.19 = 0.3534 °C

h let me think

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.