Question

Calculate the osmotic pressure at 36.6 degrees C of a solution made by dissolving 9.18 g of glucose in 34.2 mL of solution. Enter your answer using 2 decimal places!!!!

Answer 1

38.35 bar

Step-by-step explanation:

We are given that

Temperature=T=36.6 degree Celsius=36.6+273=309.6 K

Given mass of glucose=9.18 g

Molar mass of glucose(
`C_6H_(12)O_6=6(12)+12(1)+6(16)`=180 g

Mass of c=12 g,mass of hydrogen=1 g, mass of O=16 g

Volume of solution=34.2 mL

Molarity of solution=
`(given\;mass)/(molar\;mass* volume)* 1000`

Where volume (in mL)

Molarity of solution=
`(9.18)/(180* 34.2)* 1000=1.49 M`

We know that

Osmotic pressure=
`\pi=MRT`

Where M=Molarity of solution

R=Constant=0.08314 Lbar/mol k

T=Temperature in kelvin

Using the formula

`\pi=1.49* 0.08314* 309.6=38.35 bar`

Hence, the osmotic pressure=38.35 bar