Question

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

Answer 1

Answer:

$\huge\underline\mathtt\red{Answer:}$

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=

$1.9 \times 0.002 \times 0.0821 \times 300$

0.0935 atm

= 0.0935×1.013 bar

=0.094bar

Explanation:

Good Afternoon :)

Answer 2

Answer:

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=

1.9 \times 0.002 \times 0.0821 \times 3001.9×0.002×0.0821×300

0.0935 atm

= 0.0935×1.013 bar

=0.094bar