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The nonvolatile, nonelectrolyte glucose, C6H12O6 (180.2 g/mol), is soluble in water H2O. Calculate the osmotic pressure (in atm) generated when 11.9 grams of glucose are dissolved in 217 mL of a water solution at 298 K.

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Answer: The osmotic pressure of the solution is 7.44 atm

Step-by-step explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:


\pi=iMRT

or,


\pi=i* \frac{m_(solute)* 1000}{M_(solute)* V_(solution)\text{ (in mL)}}}* RT

where,


\pi = osmotic pressure of the solution = ?

i = Van't hoff factor = 1 (for non-electrolytes)


m_(solute) = given mass of glucose = 11.9 g


M_(solute) = molar mass of glucose = 180.2 g/mol


V_(solution) = Volume of solution = 217 mL

R = Gas constant =
0.0821\text{ L atm }mol^(-1)K^(-1)

T = temperature of the solution = 298 K

Putting values in above equation, we get:


\pi=1* (11.9* 1000)/(180.2* 217)* 0.0821\text{ L. atm }mol^(-1)K^(-1)* 298K\\\\\pi=7.44atm

Hence, the osmotic pressure of the solution is 7.44 atm

User Marco Fantasia
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