Answer:
The reaction is exothermic.
Explanation:
.3 liters... im pretty sure this is correct!!
The given question is incomplete. The complete question is:
When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.
Answer: The vant hoff factor for sodium chloride in X is 1.9
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant
i = vant hoff factor = 1 ( for non electrolyte)
m= molality =

Now Depression in freezing point for sodium chloride is given by:
= Depression in freezing point
= freezing point constant
m= molality =


Thus vant hoff factor for sodium chloride in X is 1.9
Ba2Cl
NaS2
The numbers are in subscript
The number of molecules : 4.967 x 10²⁴
<h3>Further explanation
</h3>
A mole is a number of particles(atoms, molecules, ions) in a substance
This refers to the atomic total of the 12 gr C-12 which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles
Can be formulated :
N = n x No
N = number of particles
n = mol
No = 6.02.10²³ = Avogadro's number
8.25 moles of C₈H₁₈
The number of molecules :
