[tex]\small\fbox\pink{Good Morning}[/tex] 0.002 molar solution of Nacl having DOD of 90% at 27°C had osmotic pressure?​

2 thoughts on “[tex]\small\fbox\pink{Good Morning}[/tex] 0.002 molar solution of Nacl having DOD of 90% at 27°C had osmotic pressure?​”

1. Answer:

   [tex]\huge\underline\mathtt\red{Answer:}[/tex]

   M= 0.002

   T= 27° celcius (300 in kelvin)

   Degree of dep= 90%= 0.9

   For Nacl= Degree of dep+1

   0.9+1= 1.9

   Osmotic Pressure=

   [tex]1.9 \times 0.002 \times 0.0821 \times 300[/tex]

   0.0935 atm

   = 0.0935×1.013 bar

   =0.094bar

   Explanation:

   Good Afternoon 🙂

   Reply

2. [tex]\small\fbox\pink{Answer}[/tex]

   M=0.002

   T=27°C = 300K

   Degree of dep = 90% = 90/100 = 0.9

   For NaCl, i = dod + 1

   =0.9 +1

   = 1.9

   Osmotic pressure= icst

   =1.9 × 0.002× 0.0821× 300

   =0.0935 atm

   =0.0935 × 1.013 bar

   = 0.094 bar

   Reply