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Chapter 11: Problem 24

Calculate the osmotic pressure of each of the following aqueous solutions at \(27^{\circ} \mathrm{C}\) a. \(10.0 \mathrm{g}\) of \(\mathrm{NaCl}\) in \(1.50 \mathrm{L}\) of solution b. \(10.0 \mathrm{mg} / \mathrm{L}\) of \(\mathrm{LiNO}_{3}\) c. \(0.222 M\) glucose d. \(0.00764 M \mathrm{K}_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
Question: Calculate the osmotic pressure for each of the following aqueous solutions at 27掳C: a) 10.0 g NaCl in 1.50 L of solution, b) 10.0 mg/L LiNO鈧�, c) 0.222 M glucose, and d) 0.00764 M K鈧係O鈧�. Answer: a) The osmotic pressure for the 10.0 g NaCl in 1.50 L of solution is approximately 5.60 atm. b) The osmotic pressure for the 10.0 mg/L LiNO鈧� solution is approximately 0.00714 atm. c) The osmotic pressure for the 0.222 M glucose solution is approximately 5.46 atm. d) The osmotic pressure for the 0.00764 M K鈧係O鈧� solution is approximately 0.562 atm.

Step by step solution

01

a. 10.0 g NaCl in 1.50 L of solution

First, determine the number of moles of NaCl and find the molarity. To calculate moles of NaCl: moles = mass (g) / molar mass (g/mol) NaCl molar mass: 58.44 g/mol moles = 10.0 g / 58.44 g/mol 鈮� 0.1712 mol To calculate molarity (M): M = moles / volume M = 0.1712 mol / 1.50 L 鈮� 0.1141 M For NaCl, the value of van 't Hoff factor (i) is 2, since it dissociates into two ions (Na\(^+\) and Cl\(^鈭抃)). Now, use the formula for osmotic pressure: Osmotic Pressure = iMRT R = 0.0821 L atm/mol K T = 27掳C + 273.15 = 300.15 K Osmotic Pressure = 2 脳 0.1141 M 脳 0.0821 L atm/mol K 脳 300.15 K 鈮� 5.60 atm
02

b. 10.0 mg/L LiNO鈧�

First, determine the number of moles of LiNO鈧� and find the molarity. To calculate moles of LiNO鈧�: moles = mass (g) / molar mass (g/mol) LiNO鈧� molar mass: 68.94 g/mol (approximately) moles = 10.0 mg 脳 10\(^{-3}\) / 68.94 g/mol 鈮� 1.45 脳 10\(^{-4}\) mol To calculate molarity (M): M = moles / volume M = 1.45 脳 10\(^{-4}\) mol / 1 L = 1.45 脳 10\(^{-4}\) M For LiNO鈧�, the value of van 't Hoff factor (i) is 2, since it dissociates into two ions (Li\(^+\) and NO鈧僜(^鈭抃)). Now, use the formula for osmotic pressure: Osmotic Pressure = iMRT Osmotic Pressure = 2 脳 1.45 脳 10\(^{-4}\) M 脳 0.0821 L atm/mol K 脳 300.15 K 鈮� 0.00714 atm
03

c. 0.222 M glucose

Glucose (C鈧咹鈧佲倐O鈧�) is a non-electrolyte and does not dissociate in aqueous solution, so the van 't Hoff factor (i) for glucose is 1. Now, use the formula for osmotic pressure: Osmotic Pressure = iMRT Osmotic Pressure = 1 脳 0.222 M 脳 0.0821 L atm/mol K 脳 300.15 K 鈮� 5.46 atm
04

d. 0.00764 M K鈧係O鈧�

First, determine the van 't Hoff factor (i) for K鈧係O鈧�. K鈧係O鈧� dissociates into 3 ions: 2K\(^+\) and 1SO\(_4^{2-}\), so the van 't Hoff factor (i) is 3. Now, use the formula for osmotic pressure: Osmotic Pressure = iMRT Osmotic Pressure = 3 脳 0.00764 M 脳 0.0821 L atm/mol K 脳 300.15 K 鈮� 0.562 atm

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