Given,
Molality = .5M
We know, Kf = 1.86
Now,
Δ T = M x Kf
= 1.86 * 0.5
= 0.93 °C
Now,
Freezing point = Initial freezing point - Δ T
= 0°C - 0.93°C
= -0.93°C
So, the answer is A
Ethanol is a polar molecule containing hydrogen bonds at the OH groups. These bonds are much stronger and therefore take much more energy to break, hence the higher melting and boiling points
Ethane on the other hand is non-polar as it is a hydrocarbon and only contains London Dispersion forces between molecules, which are the weakest of the intermolecular forces, meaning it is easy to break them.
Answer:
V/V% = 8.2%
Explanation:
Given data:
Volume of methanol = 37.5 mL
Volume of solution = 456 mL
V/V% = ?
Solution:
V/V% = [volume of solute / volume of solution ]×100
V/V% = 37.5 mL / 456 mL × 100
V/V% = 0.08× 100
V/V% = 8.2%