If a chemist wishes to dilute a 1.000 x 103 mg/l stock solution to prepare 5.000 x 102 ml of a working standard that has a concentration of 6.250 mg/l, what volume of the 1.000 x 103 mg/l standard solution is needed?

Given data:

Concentration of stock solution which is the initial solution = M1 = 1.00 * 10^3 mg/L

Concentration of prepared solution i.e final solution = M2 = 6.250 mg/L

Volume of final solution, V2 = 5.000 * 10^2 ml

To determine:

The initial volume i.e. volume of the stock solution required = V1

Calculation:

Use the relation

M1V1 = M2V2

V1 = M2V2/M1 = 6.250 mg.L-1 * 5.000*10^2 ml/1.00 * 10^3 mg.L-1

V1 = 3.125 *10^3 ml

The volume of stock solution required is 3.125 *10^3 ml

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.