How many moles in 1.00g of salicylic acid?

-  0.00724 moles.

Solution:- Grams of salicylic acid are given and it asks to calculate the moles.

For the moles, we divide the grams by the molar mass.

Formula of salicylic acid is

Molar mass = 7(12.01)+6(1.01)+3(16.00)

= 84.07+6.06+48.00

= 138.13 gram per mol

We could make the dimensional analysis set up for showing unit cancellation and have the moles.

= 0.00724 moles

So, there will be 0.00724 moles of salicylic acid in it's 1.00 gram.

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

level 4 to level 2 : ))