In each pair, identify the solution that will have a higher boiling point. explain. 1. 1.50 moles of lioh (strong electrolyte) and 3.00 moles of koh (strong electrolyte) each in 1.0 kg of water 2. 0.40 mole of al(no3)3 (strong electrolyte) and 0.40 mole of cscl (strong electrolyte) each in 1.0 kg of water

- (1) 3.00 moles of KOH in 1.0 kg of water (2) 0.40 mole of in 1.0 kg of water.

Explanations:- Elevation in boiling point is directly proportional to the molality of the solution.  The equation used for this is written as:

where, is the elevation in boiling point, i is the Van't hoff factor, m is the molality of the solution and is the molal elevation constant for the solvent used.

Van't hoff factor is the theoretical number of ions an ionic compound give when it breaks. For example NaCl breaks to give and . So, the value of i for NaCl is 2.

In the first pair we have LiOH and KOH. LiOH  gives lithium ion and hydroxide ion, so the value of i for this is 2. Similarly, KOH gives potassium ion and hydroxide ion and the value of i for this is also 2.

So, for this pair the deciding parameter is the molality. Molality of KOH is higher than LiOH and so the boiling point will be higher for KOH.

For second pair, the molality is same for both the solutions so the deciding parameter here is the value of van't hoff factor. CsCl breaks to give cesium ion and chloride ion and so the value of i for this is 2. breaks to give one ion and three ions. So, the value of i for this is 4.

Since the value of i is higher for aluminium nitrate, it's boiling point will be higher.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.