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Answer: The osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.
Explanation:
Given: Mass = 5.0 g
Volume = 1 L
Molar mass of sucrose = 342.3 g/mol
Moles are the mass of a substance divided by its molar mass. So, moles of sucrose are calculated as follows.
[tex]Moles = \frac{mass}{molarmass}\\= \frac{5.0 g}{342.3 g/mol}\\= 0.0146 mol[/tex]
Hence, concentration of sucrose is calculated as follows.
[tex]Concentration = \frac{moles}{Volume (in L)}\\= \frac{0.0146 mol}{1 L}\\= 0.0146 M[/tex]
Formula used to calculate osmotic pressure is as follows.
[tex]\pi = CRT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure
C = concentration
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]\pi = CRT\\= 0.0146 \times 0.0821 L atm/mol K \times 298 K\\= 0.357 atm (1 atm = 760 torr)\\= 271.32 torr[/tex]
Thus, we can conclude that the osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.
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