For each of these solutions, determine the osmotic pressure of thesolution at ST

Question

For each of these solutions, determine the osmotic pressure of thesolution at STP. (Assume these solutions behave ideally.) (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 (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

Navigate

• Browse (All)
• Browse F
• Subjects

---