Suppose you are studying the Ksp of CaCl2, which has a molar mass of 110.98 g/mol, at multiple temperatures. You dissolve 4.99 g of CaCl2 in 10.0 mL of water at 100 oC and cool the solution. At 90 oC, a solid begins to appear. What is the Ksp of CaCl2 at 90 oC

Hence the Solubility product,  

Ksp = [Ca2+] [Cl-]2  

or, Ksp = (4.5) (9)2  

or, Ksp = 364.5

Explanation:  

Mass of CaCl2 = 4.99 g  

Molar mass of CaCl2 = 110.98 g/mol  

Moles of CaCl2  

= given mass/ molar mass  

= 4.99/ 110.98  

= 0.045  

Volume = 10.0 mL = 0.01 L  

CaCl2 dissociates into its ion as:  

CaCl2 (s)  \rightleftharpoons Ca2+ (aq) + 2 Cl- (aq)  

At 90°C, the solution is saturated with Ca2+ and Cl- ions.

Moles of Ca2+ = Moles of CaCl2 dissolved = 0.045  

Moles of Cl- = 2 x ( Moles of CaCl2 dissolved) = 2 x 0.045 = 0.09

[Ca2+] = Moles/ Volume = 0.045/ 0.01 = 4.5 M  

[Cl-] = 0.09/ 0.01 = 9 M  

Solubility product,  

Ksp = [Ca2+] [Cl-]2  

or, Ksp = (4.5) (9)2  

or, Ksp = 364.5

i don t now

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