Question

Question 21 of 52 Submit What is the osmotic pressure of a solution made from 22.3 g of methanol (MM = 32.04 g/mol) that was added to water to make 1241 mL of solution at 25.0 °C? atm (1 2 3 C +/- x 100

Answer 1

Osmotic pressure is given by Т - НА Т. l li - Van't Hoff indea - Temperature (K) Cras constant Since CH₂OH is non electrolytel, hence i=61, Mod and 4 - Holk = a + — Moles of Methanol = 22.3 - a 0.696 Moly 32:04 - hence, molarity = 0.696 M = 0.56 M (1241 hence I = (0.56) * (0.08206) (298) XI - R 0.082 Latry Molk. - π = 0.0459 atm.

Answer 2

SOLUTION :

Formula for osmotic pressure is :

Π = I * C * R * T

where, 

Π = osmotic pressure (atm)

i = Van’t Hoff Factor = 1 for methanol being non-electrolyte.

C = molar concentration of the solute in the solution (mol/L)

R = Gas constant = 0.0821 atm . L . mol^-1 . K^-1 

T = Temperature in ºK.

Number of mols. Of methanol (CH3OH) 

= Weight of methanol / Mol. Mass

= 22.3 g / 32.04 g . mol^-1

= 0.696 mol.

Volume of solution = 1241 mL = 1.241 L

So, C = 0.696/1.241 = 0.56084

T = 25 + 273 = 298ºK

Hence,

Π = 1 * 0.56084 * 0.0821 * 298 = 13.7214 atm

The osmotic pressure for the  given case is = 13.7214 atm (ANSWER)