The solution would be like this for this specific problem:
For freezing point
Tf, soln= -114.6-(i*Kf*m)
Tf, soln= -114.6 - (1*1.99*0.40)
Tf= -115.4 C
For boiling point:
Tb=78.4+(i*Kf*m)
Tb=78.4+(1*1.99*.40)
Tb= 79.2 C
So the boiling and freezing point of a 0.40m solution of sucrose in ethanol is 79.2 C and -115.4 C respectively.
Answer:
b. ICl experiences dipole-dipole interactions
Explanation:
The molecule with the <em>stronger intermolecular forces</em> will have the higher boiling point.
In Br₂, the Br-Br bond has a <em>dipole moment of zero</em>, because the two atoms are identical.
In ICl, the I-Cl bond, has two different atoms. One must be more electronegative than the other, so there will be a <em>non-zero bond dipole</em>.
ICl will have the higher boiling point.
a is <em>wrong</em>. Br₂ is nonpolar, so it has no dipole-dipole interactions.
c is <em>wrong</em>. Br₂ cannot form hydrogen bonds, because there is no hydrogen.
d is <em>wrong</em>. ICl has dipole-dipole interactions.
1 becuse its all it can hold
