Answer: The value of
for chloroform is
when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
Explanation:
Given: Moles of solute = 0.793 mol
Mass of solvent = 0.758

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

Now, the values of
is calculated as follows.

where,
i = Van't Hoff factor = 1 (for chloroform)
m = molality
= molal boiling point elevation constant
Substitute the values into above formula as follows.

Thus, we can conclude that the value of
for chloroform is
when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
Answer:
what are we doing there???
bye, have a nice day!
the big number describes the number ratio in a chemical equation
so for example,
2H2 + O2 --> 2H2O means
2 moles of hydrogen reacts with one mole of oxygen to form 2 moles of water
and as you know, the small (subscript) number determines the number of atoms of that element in one molecule of a compound
so I believe that drawing a normal lewis structure ( O=O ) should be correct
Aluminum foil I think but if not goodluck :)