calculate the freezing poing depression for a solution prepared by dissolving 25
ID: 590289 • Letter: C
Question
calculate the freezing poing depression for a solution prepared by dissolving 25.0g of octane in 175g of benzane calculate the freezing poing depression for a solution prepared by dissolving 25.0g of octane in 175g of benzane an initial mixture of N2 and H2 gases that can be 10. represented as follows: The gases react to form ammonia gas (NH3) as represented by the following concentration profile: Time a. Label each plot on the graph as N2, H2, or NH, and explain your answers. D. Explain the relative shapes of the plots. he n is equilibrium reached? How do you know?Explanation / Answer
You need to have molal depression in freezing point constant Kf = 4.3 oC/m obtained from internet(please verify it with the given value in your problem).
T = i Kf m
T = depression in freezing point
i = vanthoff factor ( used when there is ionization of solute) our solute is non ionizable and hence i = 1
Kf = freezing point deprsssion constant
m = molality of solute
molar mass of octane = 114.23 g/mol
moles of octane = mass / molar mass = 25g / 114.23 g/mol = 0.2188 mol
mass of solvent = 175g = 0.175 kg
molality = moles of solute / mass of solvent in kg = 0.2188 mol / 0.175 kg = 1.25 m
T = 1 x 4.3 oC/m x 1.25 m = 5.38 oC
Ans = 5.38 oC
N2 + 3H2 -----> 2NH3
concentration of hydrogen should be highest and decreasing over the time to form ammonia and hence it is represented by green line
nitrogen is less than hydrogen but should be decreasing over the course of reaction and hence can be represented by blue line
ammonia concentration should start from zero and increasing, represented by red
equilibrium is reached when all of them have constant concentration,ie. when the lines become parallel to x axis
as y axis represents concentration
comment if you need more help
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