The following v_x occur at 500 vc. Arranging them in order of inverting tendency
ID: 997517 • Letter: T
Question
The following v_x occur at 500 vc. Arranging them in order of inverting tendency to process to completion (least completion rightarrow greatest completion) 2NOCl 2NO + cl_2 k_p = 1.7 Times 10^-2 N_2O_4 2NO_2 k_p = 1.5 Times 10^-3 2SO_3 2SO_2 + O_2 k_p = 1.3 Times 10^-5 k_p = Times 10^-5 What is the osmotic pressure of a 0.25 M solution of sucrose at 37 degree C? (R = 0.082 L atm/K-mol) 6.6 Times 10^-5 atm 0.76 atm 6.4 atm 100 atm 940 atm Which of the following aqueous solution has the lowest freezing point? 0.25 m K.C1 0.l5mNa_2SO_4 0.12 m Ca(NO_3)_2 pure water 0.20 m C_2H_6O_2 (ethylene glycol)Explanation / Answer
Given data:
Concentration of sucrose solution C = 0.25 M
T = 37 oC = 37 + 273.15 = 310.15 k
R = 0.0821 L.atm.mol-1K-1.
Osmatic pressure () = ? atm.
Formula:
Osmatic pressure in atmosphere is related with C, R, T as
= CRT
= 0.25 x 0.0821 x 310.15
= 6.37 atm
Hence osmatic pressure of 0.25 M sucrose solution at 37 oC is 6.37 atm.
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9) Depression in freezing point is a colligative property.
Depression in freezing point of solution depends on the concentration of solute.
In case of ionic solute which dissociates into the number of ions in the solution expected to be showing more depression in freezing property.
A) 0.25 KCl dissolves in solvent to give 2 ions K+ and Cl– hence total ionic concentration will be 0.25 + 0.25 = 0.5 M
B) Na2SO4 ionizes into 2 Na+ and 1 SO42- ions i.e. 3 ions. Hence on dissolution 0.15 M NaSO4 ionic concentration will be 3 x 0.15 = 0.45 M
C) Ca(NO3)2 ionizes into 1 Ca2+ and 2 NO3– ions i.e. 3 ions. Hence on dissolution 0.12 M Ca(NO3)2 will give ionic concentration 3 x 0.12 = 0.36 M.
E) 0.20 M ethylene glycol. Ethylene glycol will not ionize in the aqueous solution and hence solute concentration is 0.20M only.
Among all these values ionic concentration is highest for 0.25 M KCl and hence depression in freezing point in 0.25 KCl solution will be most.
Ans: (A)
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