Question

what is the osmotic pressure at 20 C° of a solution made by dissolving 0.69 grams of sucrose, C12H22O11, in 100 mL of water?​

Answer 1

The osmotic pressure is 0.4808 atm

Step-by-step explanation:

Given:

Temperature =
`20^(\circ) Celsius`

Volume of sucrose = 0.69 grams

volume of water = 100 mL

To Find:

The osmotic pressure =?

Solution:

Step 1: lets find the number of moles in 0.69 grams of sucrose(
`C_(12)H_(22)O_(11))`

number of moles in 0.69 grams of sucrose =
`(1 mol)/(342 grams) * 0.69 g`

=>
`(1 mol)/(342 grams) * 0.69 g`

=>
`0.0029 * 0.69 g`

=>0.002 moles

Step2 :Dividing number of moles by the amount of litres in the solution

=>
`(0.002)/(0.100)` [ 100 mL = 0.100L]

=> 0.02 M

Step 3: Converting the temperature to kelvin

=>20 +273

=>293 K

Step 4: Finding the value of osmotic pressure

`\text{Osmotic pressure} \pi = MRT`

where

M is the Molarity

R is gas constant

T is the temperature

Now substituting the values,

`\text{Osmotic pressure} \pi = 0.02 * 0.08206 \ times 293`

`\text{Osmotic pressure} \pi = 0.4808 atm`