A4.55 l sample of water contains 0.115 g of sodium ions. determine the concentration of sodium ions in ppm if the density of the solution is 1.00 g/ml.

25.3 ppm

Explanation: ppm stands for parts per million. It's a way for representing concentrations when a very small amount of solute is dissolved in the solvent.

1ppm =

From given information, 0.115 grams of sodium ions(solute) dissolved in 4.55 L sample of water.

Let's convert the grams of solute to mg.

1 g = 1000 mg

= 115 mg

Let's divide the mg of solute by liters of solution to get the ppm concentration.

= 25.3 ppm

(Note- mass of solute is very much negligible as compared to the mass of solution, so mass of solute is almost same as the mass of solution)

So, the concentration of sodium ions in ppm will be 25.3 .

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