Write the equation needed to solve the following problems. (Do not attempt to so

Question

Write the equation needed to solve the following problems. (Do not attempt to solve for a value, I deliberately didn't give you the information for that. Pretend the blanks have numbers; show me what equation you would use with them.) At degree C, what is the osmotic pressure of a homogeneous solution consisting of ___ grams urea (CON_2H_4) diluted with water to ____ L. Determine the vapor pressure of a solution at ___ degree C hat contains ___ g of glucose (C_6H_12O_6 in __ g of water. The vapor pressure of pure water at that temperature is ____ torr. The vapor pressure of water is 1.0 atm at 373 K, and the enthalpy of vaporization is 40.7 kJ mol^-1. Estimate the vapor pressure at temperature ___. Calculate the amount of carbon dioxide dissolved in one litre of soda pop if the manufacturer uses a pressure of 2.4 atm of CO_2 to carbonate the soda pop. The molar constant is ___ atm/M.

Explanation / Answer

1. Moles urea = 8,65 / 60,0554 = 0,1440336
M = 0,1440336/ 1,50 = 0,0960224

Osmotic Pressure = 0,0960224 x 0,08206 x 298 K = 2,35 ATM.-

2.se raoult's law for this problem

vapor pressure of solution = mole fraction of solvent * vapor pressure of pure solvent

76.6 / (6 * 12.01 + 12 * 1.01 + 6 * 16) = 0.425 moles glucose
250 / (2 * 1.01 + 16) = 13.87 moles water

mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
mole fraction of solvent = 13.87 / (13.87 + 0.425) = 0.970

vapor pressure of solution = 0.970 * 23.8 = 23.09 torr

3.
Using the Clausius-Clapeyron equation, we have:

P363 = 1.0 exp (- (40700/8.3145)(1/363 - 1/373)
= 0.697 atm

P383 = 1.0 exp (- (40700/8.3145)(1/383 - 1/373)
= 1.409 atm

Note that the increase in vapor pressure from 363 K to 373 K is 0.303 atm, but the increase from 373 to 383 K is 0.409 atm. The increase in vapor pressure is not a linear process

4.When a gas is dissolved in a liquid, the concentrations will eventually reach equilibrium between the source of the gas and the solution. Henry's Law shows the concentration of a solute gas in a solution is directly proportional to the partial pressure of the gas over the solution.

P = KHC where

P is the partial pressure of the gas above the solution
KH is the Henry's Law constant for the solution
C is the concentration of the dissolved gas in solution

C = P/KH
C = 2.4 atm/29.76 atm/(mol/L)
C = 0.08 mol/L

since we only have 1 L of water, we have 0.08 mol of CO2.

Convert moles to grams

mass of 1 mol of CO2 = 12+(16x2) = 12+32 = 44 g

g of CO2 = mol CO2 x (44 g/mol)
g of CO2 = 8.06 x 10-2 mol x 44 g/mol
g of CO2 = 3.52 g

NOTE :- As values are not given, I have assumed value on my own

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