Question

1-A compound has a strong absorption at 5.59x10^16 Hz. what is the Ymax( in angstroms) for this compound? (Round answer to one decimal place)

2-A student notice that the molar absorptivity was 619 mL/(mg*m). What is the molar absorptivity in units of L/(mol*cm)? The molar mass of the compound is 210g/mol

Answer 1

Answer 2

1. To find Ymax (in Å) for a compound with a strong absorption at 5.59x10^16 Hz, we can use the following equation:

Ymax = (c * h) / (E * 10^10)

where c is the speed of light (3.00 x 10^8 m/s), h is Planck's constant (6.626 x 10^-34 J s), E is the frequency of the absorption (5.59 x 10^16 Hz), and 10^10 is used to convert from meters to angstroms.

Substituting the given values, we get:

Ymax = (3.00 x 10^8 m/s * 6.626 x 10^-34 J s) / (5.59 x 10^16 Hz * 10^10)

Ymax ≈ 5.64 Å

Therefore, the Ymax for this compound is approximately 5.64 Å.

1. The molar absorptivity (ε) relates the absorbance (A) of a solution to its concentration (c) and path length (l) according to the Beer-Lambert law:

A = ε * c * l

where c is in units of mol/L and l is in units of cm. To convert from mL/(mgm) to L/(molcm), we need to use the molar mass (M) of the compound to convert from mg to mol:

M = 210 g/mol = 210,000 mg/mol

Then we can use the following equation:

ε (L/(molcm)) = ε (mL/(mgm)) * (1/1000) * (1/M) * (100)

where the first factor converts mL to L, the second factor converts mg to mol, and the third factor converts cm to m, and the fourth factor multiplies by 100 to convert from mL/(mgm) to L/(molcm).

Substituting the given value for ε, we get:

ε (L/(molcm)) = 619 mL/(mgm) * (1/1000) * (1/210,000 mol/mg) * (100)

ε (L/(molcm)) ≈ 0.147 L/(molcm)

Therefore, the molar absorptivity of the compound is approximately 0.147 L/(mol*cm).