200 ml of an aqueous solution contains 18 g
of glucose. The osmotic pressure of the solution
at 27°C will be (Molar mass of glucose = 180 g
mol-1)
Answers
The osmotic pressure of the solution at 27°C will be 12.315 atm.
Explanation:
The volume of the aqueous solution, V = 200 ml = 0.2 L
Temperature, T = 27℃ = 27+273 = 300 K
Mass of glucose, m = 18 g
Molar mass of glucose = 180 g/mol
∴ No. of moles of glucose = 18/180 = 1/10 = 0.1 mol
The formula for calculating the osmotic pressure is given as,
π = c*R*T
where
c = concentration of the dissolved compound = n/V
R = Universal Gas Constant = 0.0821 L atm K⁻¹mol⁻¹
Now, substituting the given values in the formula, we get
π = [n/V]*R*T
⇒ π = [0.1/0.2] * 0.0821 * 300
⇒ π = 0.5 * 0.0821 * 300
⇒ π = 12.315 atm
Thus, the osmotic pressure of the solution is 12.315 atm .
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The osmotic pressure of the solution at 27 degree Celsius will be =12.315 atm
Explanation:
Volume of solution=200 ml=0.2 L
(1 L=1000mL)
Given mass of solute( glucose)=18 g
Temperature=T=27 degree Celsius=
Molar mass of glucose=180 g/mol
Number of moles of solute =
Where =Given mass of solute
=Molar mass of solute
Number of moles of glucose=
We know that
Where =Osmotic pressure
n=Number of moles of solute
R=Universal gas constant =0.0821 L atm/kmol
T=Temperature
V=Volume of solution
Substitute the value in the formula then, we get
Hence, osmotic pressure of the solution=
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