# The osmotic pressure of the solution obtained by mixing 200cm³ of 25% (mass-volume) solution of urea with 200cm³ of 3.42% solution of sucrose at 20°C is 1) 4 bar 2) 1.2 bar 3) 5.2 bar 4) 15.4 bar

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**Answer:**

(b) 1.2 Bar Hope u like my Answer and Mark me as Brainliest.

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**Explanation:**

**Explanation:**__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/mol__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/mol__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293K__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293K__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we have__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we have__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = __

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute =__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 60__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 60__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 __

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + __

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 +__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 342__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 342__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 __

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = __

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 =__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 400__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 400__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 __

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 __

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M∴ π=0.108×0.082×293__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M∴ π=0.108×0.082×293__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M∴ π=0.108×0.082×293 =2.594 atm__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M∴ π=0.108×0.082×293 =2.594 atm__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M∴ π=0.108×0.082×293 =2.594 atmwe know that 1 atm =1.01325 bar__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M∴ π=0.108×0.082×293 =2.594 atmwe know that 1 atm =1.01325 bar__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M∴ π=0.108×0.082×293 =2.594 atmwe know that 1 atm =1.01325 barSo, π=2.594×1.01325≈2.60 bar__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M∴ π=0.108×0.082×293 =2.594 atmwe know that 1 atm =1.01325 barSo, π=2.594×1.01325≈2.60 bar__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M∴ π=0.108×0.082×293 =2.594 atmwe know that 1 atm =1.01325 barSo, π=2.594×1.01325≈2.60 barHence, the correct option is D.__

__Explanation:π= Osmotic pressure =CRT R=0.082 Latm/K/molC= Molar concentration T=20+273=293KIn 400ml of solution we haveSolute = 602 + 3423.42 Molar mass of urea =60 =0.043 moles Sucrose =342 = 4000.043×1000 =0.108M∴ π=0.108×0.082×293 =2.594 atmwe know that 1 atm =1.01325 barSo, π=2.594×1.01325≈2.60 barHence, the correct option is D.__

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