For each of these solutions, determine the osmotic pressure of thesolution at ST
ID: 675937 • Letter: F
Question
For each of these solutions, determine the osmotic pressure of thesolution at STP. (Assume these solutions behave ideally.) (a) 0.213 MFeCl2 in water1 atm
(b) 0.045 m CoCl2 inwater (Assume the density of the solution is 1.0 g/mL.)
2 atm
(c) 0.123 mol fraction glucose(C6H12O6) in water (Assume thedensity of the solution is 1.02 g/mL.)
3 atm
(d) 0.0930 mol fraction NaCl in water(Assume the density of the solution is 1.00 g/mL.)
4 atm (a) 0.213 MFeCl2 in water
1 atm
(b) 0.045 m CoCl2 inwater (Assume the density of the solution is 1.0 g/mL.)
2 atm
(c) 0.123 mol fraction glucose(C6H12O6) in water (Assume thedensity of the solution is 1.02 g/mL.)
3 atm
(d) 0.0930 mol fraction NaCl in water(Assume the density of the solution is 1.00 g/mL.)
4 atm
Explanation / Answer
We Know that : a. = iCRT FeCl2 ------> Fe+2 + 2 Cl- = 3 x 0.213M x0.0821 atm-L / mole -K x 273 K = 14.322 atm b. M = m x density / 1 + molecularmass of solute / 1000) = 0.045 m x 1.0 g / ml x 1000 / 1 +183.4 = 0.244 M = iCRT CoCl2 ------> Co+2 + 2Cl- =3 x 0.244M x 0.0821 atm-L / mol-K x 273 K = 16.40 atm c. XC6H12O6 = 0.123 XH2O = 1 - 0.123 = 0.877 MC6H12O6 = X C6H12O6 / XH2O * 1 / density of thesolution = 0.123 / 0.877 * 1 / 1.02 g / ml = 0.1375 M C6H12O6 = i CRT i = 1 as glucose is non-electrolyte = 1 * 0.1375 M * 0.0821 atm -L / mole-K * 273 K = 3.08atm. d. XNaCl = 0.0930 XH2O = 1 - 0.0930 = 0.907 MNaCl = X NaCl / XH2O * 1 / density of thesolution = 0.0930/ 0.907 * 1 / 1.0 g / ml = 0.1025 M NaCl = i CRT i = 2 as sodium chloride is an electrolyte = 2 * 0.1025 M * 0.0821 atm -L / mole-K * 273 K = 4.596 atm. a. = iCRT FeCl2 ------> Fe+2 + 2 Cl- = 3 x 0.213M x0.0821 atm-L / mole -K x 273 K = 14.322 atm b. M = m x density / 1 + molecularmass of solute / 1000) = 0.045 m x 1.0 g / ml x 1000 / 1 +183.4 = 0.244 M = iCRT CoCl2 ------> Co+2 + 2Cl- =3 x 0.244M x 0.0821 atm-L / mol-K x 273 K = 16.40 atm c. XC6H12O6 = 0.123 XH2O = 1 - 0.123 = 0.877 MC6H12O6 = X C6H12O6 / XH2O * 1 / density of thesolution = 0.123 / 0.877 * 1 / 1.02 g / ml = 0.1375 M C6H12O6 = i CRT i = 1 as glucose is non-electrolyte = 1 * 0.1375 M * 0.0821 atm -L / mole-K * 273 K = 3.08atm. d. XNaCl = 0.0930 XH2O = 1 - 0.0930 = 0.907 MNaCl = X NaCl / XH2O * 1 / density of thesolution = 0.0930/ 0.907 * 1 / 1.0 g / ml = 0.1025 M NaCl = i CRT i = 2 as sodium chloride is an electrolyte = 2 * 0.1025 M * 0.0821 atm -L / mole-K * 273 K = 4.596 atm. XH2O = 1 - 0.0930 = 0.907 MNaCl = X NaCl / XH2O * 1 / density of thesolution = 0.0930/ 0.907 * 1 / 1.0 g / ml = 0.1025 M NaCl = i CRT i = 2 as sodium chloride is an electrolyte = 2 * 0.1025 M * 0.0821 atm -L / mole-K * 273 K = 4.596 atm.Related Questions
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