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
CH_3 CH_2 Cl has two different types of hydrogen. On further chlorination they are being substituted by chlorine to give two isomeric dichlorides
The structures of the two chlorides are shown below.
Where Hb is replaced by Cl and Ha is replaced by Cl
Answer:
Lewis dot diagrams use dots arranged around the atomic symbol to represent the electrons in the outermost energy level of an atom
Explanation:
N₂O₄ (g) ⇌ 2 NO₂ (g)
Kc = [ NO₂ ]² / [ N₂O₄ ]
4.2 = [ 0.785 ] [ N₂O₄ ]
[ N₂O₄ ] = 4.2 / 0.785
[N₂O₄ ] = 5.350 M
hope this helps!