Question

(i) Find the empirical formula of a compound that contains 36% of 9gm of element A and 64% of 4gm of element B.
(ii) A quanta of light having energy E has a wavelength of 750 nm. Find wavelength of
photon corresponding to energy equal to 5E.
(iii) In three moles of ethane (C2H6), calculate the following:
a. Number of moles of carbon atoms.
b. Number of moles of hydrogen atoms.
c. Number of molecules of ethane.​

Answer 1

Answer:

1) atomic mass of Pb = 207 g/mol

atomic mass of N = 14 g/mol

atomic mass of O = 16 g/mol

percentage composition:

Pb = 62.5% , N = 8.5% and O = 29%

Now finding percentage mass ratio,

Pb ⇒ 62.5/207 = 0.3063

N ⇒ 8.5/14 = 0.6071

O ⇒ 29/16 = 1.8125

finding simple ratio of given elements,

Pb : N : O : : 0.3063 : 0.6071 : 1.8125

⇒Pb : N : O : : 0.3 : 0.6 : 1.8

⇒ Pb : N : O : : 1 : 2 : 6

hence, empirical formula of compound is PbN2O6

2)The frequency of the light is 1.25 × 10^15 s^-1

Given:

Wavelength = 7200A°

Light of energy = 3E

Solution:

The energy of the light and the wavelength of light is inversely proportional to each other.

Relationship between wavelength and frequency and wavelength is given below:

frequency = c / lambda

Where,

c = Velocity of light (m/s)

= Wavelength (1/m)

f = 3 × 3 × 10^8 / 7200× 10^-10

f = 1.25 × 10^15 s^-1

3)

1 molecule of ethane contains 2 carbon atoms 

so 3 moles of ethane contains 6 mole of carbon atoms.

1 molecule of ethane contains 6 hydrogen atoms 

so 3 moles of ethane contains 18 mole of hydrogen atoms.

1 mole of ethane contains 6.023×10²³ molecules of ethane

so 3 mole of ethane contains 3×6.023×10²³ =18.069 ×10²³ molecules of ethane

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