Calculate the freezing point of a solution made from 220g of octane (c hua), molar mass = 114,0 gmol dissolved in 1480 g of benzene. benzene freezes at 5.50"c and its kvalue is 5.12c/m. -1.16°c 0.98°c 666"c 12 2°c 5.49°c 10 12 am a a 2019 backspace yuo pill но кl

Freezing point of a solution will be

Explanation:

Depression in freezing point is given by:

= Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

= freezing point constant =

m= molality

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

Thus the freezing point of a solution will be

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.