If the total number of atoms per unit cell in an hcp structure and a bcc structure gets halved, then ratio of percentage voids in hcp and bcc structures is

In HCP ; Packing efficiency:

Atoms occupy:74%

Empty space(voids): 28% 

In BCC; Packing efficiency:

Atoms occupy: 68% 

Empty space(voids): 32% 

According to the given condition if the total number of atoms per unit cell in a hcp structure and a bcc structure get halved. Then packing efficiency in HCP and BCC respectively given as:

In HCP ; Packing efficiency:

Atoms occupy: 74/2 = 37%

Empty space(voids): (100 - 37) = 63%

In BCC; Packing efficiency:

Atoms occupy: 68/2 = 34%

Empty space(voids): (100 - 34) = 66%

Then new ratio of their voids = 63%/66% = 21/22

What is relation between mole fraction and molality of solute ?

ILLUSTRATIVE EXAMPLES:

(1):Calculate the mole fraction of solute in the 12 molal aqueous solution of CaCO3.

(2):The mole fraction of glucose in aqueous solution is 0.2 , The molality of solution will be.

(3):If the mole fraction of a solute is changed from 1/4 to 1/2 in the 800 gm of solvent . then the ratio of molality will be .

(4):Mole fraction of I2 (iodine) in benzene (C6H6) is 0.2 . Calculate the molality of iodine (I2) in benzene.

(5):Calculate the amount of water that should be added to 16 gm methanol to make mole fraction of methanol as 0.25.?

Calculate the amount of water that should be added to 16 gm methanol to make mole fraction of methanol as 0.25.?

Mole fraction of I2 (iodine) in benzene (C6H6) is 0.2 . Calculate the molality of iodine (I2) in benzene.

The relation  between molarity and mole fraction is given as below:

If the mole fraction of a solute is changed from 1/4 to 1/2 in the 800 gm of solvent . then the ratio of molality will be .