Aseries of five calibration standards 25, 50, 75 and 100 mg/l potassium is required to be made from a 1000mg/l potassium standard solution. if the final volume of the calibration standards is 100 ml, what is the required volume of the 1000 mg/l potassium standard solution?

Answer : The volume required is, 25 mL

Explanation :

First we have to determine the mass of potassium present in 100 mL or 0.1 L of each standards.

For 25 mg/L :

As, 1 L volume has mass of potassium = 25 mg

So, 0.1 L volume has mass of potassium =

For 50 mg/L :

As, 1 L volume has mass of potassium = 50 mg

So, 0.1 L volume has mass of potassium =

For 75 mg/L :

As, 1 L volume has mass of potassium = 75 mg

So, 0.1 L volume has mass of potassium =

For 100 mg/L :

As, 1 L volume has mass of potassium = 100 mg

So, 0.1 L volume has mass of potassium =

Now we have to determine the total mass of potassium.

Total mass of potassium = 2.5 + 5.0 + 7.5 + 10 = 25 mg

Now we have to determine the required volume of the 1000 mg/L Potassium standard solution.

As, 1000 mg mass of potassium present in 1 L volume

So, 25 mg mass of potassium present in volume

Volume required = 0.025 L = 25 mL     (conversion used : 1 L = 1000 mL)

Therefore, the volume required is, 25 mL

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

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e