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Answer:
The density of Ammonia : 6.25.10⁻⁴ g/cm³
Further explanation
Density is a quantity derived from the mass and volume
Density is the ratio of mass per unit volume
With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller density
The unit of density can be expressed in g/cm³ or kg/m³
Density formula:
\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}ρ = Vm
ρ = density
m = mass
v = volume
A common example is the water density of 1 gr/cm³
Ammonia has a density of 0.625 g/L, then convert to g/cm³ :
\begin{gathered}\rm 1~L=1~dm^3=10^3~cm^3\\\\0.625~\dfrac{g}{L}\times \dfrac{1~L}{10^3~cm^3}\\\\\rho=\dfrac{0.625}{10^3}}\dfrac{g}{cm^3}=\boxed{6.25.10^{-4}\dfrac{g}{cm^3}}\end{gathered}
Learn more
This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
