Adialysis bag containing a .4m of nacl solution was submerged in a beaker containing a .6m of nacl solution. the dialysis tubing is permeable to water but not salt. the temperature of both solutions is 23 degrees celsius. determine the initial solute potential for nacl solution in the beaker.

The answer is 29.48bars

Explanation:

The formula for calculating solute potential is

Ѱs = iCRT.

i= ionization constant

C= MOLAR concentration

R =Pressure constant

T= Temperature in Kelvin (K)

The beaker contains 0.6M of NaCl

NaCl osis a salt that dossolvesdoss Olveston completely in water, hence its ionization constant is 2.

Temperature in Celsius is 23°C.

To convert from Celsius to Kelvin (K) we add 273

=23+273

=296K

Pressure constant (R) is 0.0831/liter bar/ mole°K

Therefore the initial solute potential will be calculated as

Ѱs = iCRT

= 2 × 0.6 × 0.083 × 296

= 29.48bars

condensation. good luck!