A chemist prepares a solution of iron(III) chloride (FeCl3) by measuring out 0.40 g of FeCl, into a 100. mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Cl anions in the chemist's solution. Be sure your answer is rounded to the correct number of significant digits.

Molarity Cl- = 0.074 M

Explanation:

Step 1: Data given

Mass of FeCl3 = 0.40 grams

Molar mass FeCl3 = 162.2 g/mol

Volume = 100 mL = 0.100 L

Step 2: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 0.40 grams / 162.2 g/mol

Moles FeCl3 = 0.00247 moles

Step 3: Calculate molarity of FeCl3

Molarity = moles FeCl3 / volume

Molarity = 0.00247 moles / 0.100 L

Molarity = 0.0247 M

Step 4: Calculate molarity of Cl- ions

In 1 mol FeCl3 we have 3 moles Cl-

in 0.00247 moles FeCl3 we have 3*0.00247 = 0.00741 moles Cl-

Molarity Cl- = moles Cl- / volume

Molarity Cl- = 0.00741 moles / 0.100 L

Molarity Cl- = 0.074 M

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally