8. 71.67 g of InBr2 was dissolved in water to a final volume of 650 mL. (A) What is the molar concentration of indium in the resulting solution? (B) If 25.0 mL of the solution in part A was diluted to 75.0 mL, what is the molar concentration of the resulting solution?

A) 0.39 M is the molar solution of the resulting solution.

B) The resulting solution will have molarity of 0.13M

Explanation:

The atomic weight of 2 atoms of bromine = 79.904 gms

The atomic weight of 1 atom of In = 114.818

Atomic weight of InBr2 = 274.26 gm

From the data mass percentage is calculated ie. 58.19% of bromine, 41.80 % of Indium

so from the mass of the InBr2 the percentage of indium can be calculated as

41.088/100 * 71.67

= 29.447 grams

No of moles of indium = 29.04/114.818

= 0.256 moles

In the solution of 650 ml 0.256 moles of Indium are there

M=n/C

= 0.256/650ml

= 0.256/0.65 litre

= 0.39 M

B) in 25 ml of the solution the molarity is 0.39 which is diluted to 75 ml ie three fold diluted the molarity decreases.

from the formula M1V1=M2V2

= 0.39*25 = M2 * 75

M2 = O. 39*25/75

= 0.13 M