Question

5.   Millikan’s oil drop experiment demonstrated which of the following (circle all that apply):

A. Negative charge is quantized

B. The mass of an electron is comparable to that of an oil drop

C. Heavier oil drops need to contain more electrons in order to be suspended in an electric field

D. Nuclei contain protons and neutrons

E.   The charge measured for each suspended oil drop is a whole-number multiple of the charge of an electron

6. What are the mass percentages of C, H, N and Cl in [Fe(C₅H₅N)₄Cl₂]? How would these numbers change if one of the C₅H₅N groups were lost?

Answer 1

Ans 5: option is A, C, E

Reason: Mulliken oil drop experiment gives information about charge on electron which was dicovered by using oil drop.

Ans 6: The molecular weight of [Fe(C5H5N)4Cl2] is 443g/mol

The % of carbon = mass of carbon/total mass of compound = (12 x 5 x 4)/443 * 100 = 54.17%

% H = mass of hydrogen/total mass of compound = (1 x 5 x 4)/443 * 100 = 4.55 %

% Cl = mass of chlorine/total mass of compound = (35.5 x 2)/443 * 100 = 16.02 %

% N = mass of Nitrogen/total mass of compound = (14 x 4)/443 * 100 = 12.64 %

If one of the C5H5N is lost, the new % will be:

The molecular weight of [Fe(C5H5N)3Cl2] is 443g/mol

The % of carbon = mass of carbon/total mass of compound = (12 x 5 x 3)/364 * 100 = 49.45%

% H = mass of hydrogen/total mass of compound = (1 x 5 x 3)/364 * 100 = 4.15 %

% Cl = mass of chlorine/total mass of compound = (35.5 x 2)/364 * 100 = 19.50 %

% N = mass of Nitrogen/total mass of compound = (14 x 3)/443 * 100 = 11.56 %