molality of a solution containing 0.85 gram of NH3 in 100 ml liquid of density 0.85 g /cm?
Dear Student,
Given,
Density = 0.85 g/cm3 = 0.85 g/0.001L = 85 g/0.1L
Therefore,
There are 0.85 grams of NH3 in 100cm3 ,i.e, 0.85 g /0.1 L solution
Thus, Density = 0.85 g/cm3 = 0.85 g/0.001L = 85 g/0.1L
Therefore,
There are 0.85 grams of NH3 in 100cm3 ,i.e, 0.85 g /0.1 L solution
Mass of water in solution (0.1L) = 85-0.85
= 84.15 g = 0.08415kg
Now,
Number of moles of NH3 = Mass/Molecular weight of NH3
= 0.85/17 = 0.05mol
Hence,
Molality = Number of moles / mass of water in kg
= 0.05/0.08415
= 0.597 m
Regards,
= 84.15 g = 0.08415kg
Now,
Number of moles of NH3 = Mass/Molecular weight of NH3
= 0.85/17 = 0.05mol
Hence,
Molality = Number of moles / mass of water in kg
= 0.05/0.08415
= 0.597 m
Regards,