<span>1.16 moles/liter
The equation for freezing point depression in an ideal solution is
ΔTF = KF * b * i
where
ΔTF = depression in freezing point, defined as TF (pure) ⒠TF (solution). So in this case ΔTF = 2.15
KF = cryoscopic constant of the solvent (given as 1.86 âc/m)
b = molality of solute
i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1.
Solving for b, we get
ΔTF = KF * b * i
ΔTF/KF = b * i
ΔTF/(KF*i) = b
And substuting known values.
ΔTF/(KF*i) = b
2.15âc/(1.86âc/m * 1) = b
2.15/(1.86 1/m) = b
1.155913978 m = b
So the molarity of the solution is 1.16 moles/liter to 3 significant figures.</span>
Answer: yo sorry this a hard one
Explanation:
bro
<h2>Nuclear Fission and Nuclear Fusion - Option C</h2>
Nuclear fission and nuclear fusion both of these processes can provide energy. Nuclear fission is the process in which heavy nucleus splits into smaller parts. When they split into smaller particles then it releases energy.
On the other hand, nuclear fusion is the process in which small particles fuse together to form a heavy nucleus. With the formation of heavy nucleus, it also provides energy.
Therefore, both these processes release or provide energy.
