Chem 2 questions 21.An ideal gas originally at 0.85 atm and 66 degree C was allo
ID: 882102 • Letter: C
Question
Chem 2 questions
21.An ideal gas originally at 0.85 atm and 66 degree C was allowed to expand until its final volume, pressure, and temperature were 94 mL, 0.60 atm, and 45 degree C, respectively. What was its initial volume? 22. What is the osmotic pressure (in atm ) of a 1.57M aqueous solution of urea [(NH2)2C0] at 27.0 degree C? 23. In the following phase diagram, what is occurring at point B? 24. For dilute aqueous solutions in which the density of the solution is roughly equal to that of the pure solvent, the molarity of the solution is equal to its molality. Show that this statement is correct for a 0.010-Maqueous urea (NH2)2C0 solution with a density of 1g/mL (hint: assume you have 10 L of the solution)Explanation / Answer
Q21 solution
Given data
Original pressure P1 = 0 .85 atm
Original temperature T= 66 C +273 = 339K
Original volume V1 = ?
Final pressure P2 =0.60 atm
Final volume V2= 94 ml
Final temperature T2 = 45 C +273 = 318 K
Using the combined gas law formula we can calculate the original volume of the gas
P1V1/T1 = P2V2/T2
Rearranging this equation we get
V1 = P2V2*T1 / P1T2
Lets put the values in the formula
V1 = 0.60 atm * 94 ml * 339 K / 0.85 atm * 318 K
V1 = 70.7 ml
We can round it to 71 ml
So the original volume of the gas = 71 ml
Q22 solution :-
Given data
Molarity of the urea =1.57 M
Temperature = 27 C +273 = 300 K
Osmotic pressure = ?
Formula to calculate osmotic pressure is as follows
= MRT
Where , T= temperature , M= molarity
R= 0.08206 L atm per K. mol
Lets put the values in the formula
= 1.57 M * 0.08206 L atm per K mol * 300 K
= 38.7 atm
Therefore osmotic pressure of the solution = 38.7 atm.
Q23 Solution in the phase diagram at the point B supercritical liquid is forming.
Q24 solution
Given Urea solution is 0.01 M
Lets assume we have 1.0 L solution and its density is 1 g/ml
0.01 moles of urea are present in the solution therefore
0.01 mol urea * 60.06 g per mol = 0.606 g urea
therefore if we find the total mass of solution then its 1000 g + 0.60 g = 1000.6 g
1000.6 g * 1 kg / 1000 g = 1.0006 kg
Molality = moles / kg solvent =0.01 mol / 1.0 kg = 0.010 m
Therefore molarity and molality of the dilute solution is same.
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