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
Potassium is an alkali metal with the chemical symbol K. It has an atomic number of 19, meaning that it has 19 positively charged protons. It also contains 19 electrons, which have a negative charge, and 20 neutrons, which do not hold a charge
Hope this help
The volume of a 6.67 m nacl solution
that contains 3.12 mol nacl is 0.468.
You just need to follow this formula:
C = n / V
So, it will be
V = n / C
<span> = 3.12 / 6.67 moles / Liter </span>
<span> = 0.468 L</span>
We can use combined gas laws to solve for the volume of the gas
where P - pressure, V - volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - temperature in Kelvin - 20 °C + 273 = 293 K
T2 - 40 °C + 273 = 313 K
substituting the values
V = 17.8 L
volume of the gas is 17.8 L
