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

Answer:

38.35 bar

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=$\frac{given\;mass}{molar\;mass\times volume}\times 1000$

Where volume (in mL)

Molarity of solution=$\frac{9.18}{180\times 34.2}\times 1000=1.49 M$

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

$\pi=1.49\times 0.08314\times 309.6=38.35 bar$

Hence, the osmotic pressure=38.35 bar