You need to make an aqueous solution of 0.182 M aluminum sulfate for an experiment in lab, using a 250 mL volumetric flask. How much solid aluminum sulfate should you add

Hence, 15.99 g of solid Aluminum Sulfate should be added in 250 mL of Volumetric flask.

Explanation:

To make 0.187 M of Aluminum Sulfate solution in a 250 mL (0.250 L) Volumetric flask  

The molar mass of Aluminum Sulfate = 342.15 g/mol  

Using the molarity formula:-  

Molarity = Number of moles/Volume of solution in a liter  

Number of moles = Given weight/ molar mass  

Molarity = (Given weight/ molar mass)/Volume of solution in liter  

0.187 M = (Given weight/342.15 g/mol)/0.250 L  

Given weight = 15.99 g  

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