Question 1 E. 0.0556 atm A solution containing 7.24 g of a nonvolatile compound

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Question 1

E. 0.0556 atm

A solution containing 7.24 g of a nonvolatile compound dissolved in 115.3 g of benzene has a freezing point of 3.55°C. Given that the freezing point of pure benzene is 5.45°C and the molal freezing point depression constant for benzene, Kf, is 5.08°C/m, what is the molar mass of the solute?

E. 168 g/mol

The osmotic pressure of a solution of 140 mg of an unknown protein dissolved in 5.00 mL of solvent is 0.0236 atm at 25°C. What is the molar mass of the protein?

Question 1

What is the osmotic pressure exerted at 25°C by 5.00 mL of a solution containing 224 mg of a protein with a molar mass of 19,700 g/mol? A. 0.561 atm B. 9.19 atm C. 5.62 atm D. 4.66 × 10-3 atm

E. 0.0556 atm

A solution containing 7.24 g of a nonvolatile compound dissolved in 115.3 g of benzene has a freezing point of 3.55°C. Given that the freezing point of pure benzene is 5.45°C and the molal freezing point depression constant for benzene, Kf, is 5.08°C/m, what is the molar mass of the solute?

A. 312 g/mol B. 19.4 g/mol C. 80.6 g/mol D. 23.5 g/mol

E. 168 g/mol

The osmotic pressure of a solution of 140 mg of an unknown protein dissolved in 5.00 mL of solvent is 0.0236 atm at 25°C. What is the molar mass of the protein?

A. 48,500 g/mol B. 29,000 g/mol C. 29,400 g/mol D. 725 g/mol E. 97,400 g/mol

Explanation / Answer

1)
concentration,
C = mass / (molar mass * volume in L)
= 0.224 g / (19700 g/mol* 0.005 L)
= 2.274*10^-3 L

T = 25 oC = (25 + 273) K = 298 K

P = C*R*T
= 2.274*10^-3 * 0.0821*298
= 0.0556 atm

Answer: E

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