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: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=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 = 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 + 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 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 = 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 =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 =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 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.
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