A solution is prepared by dissolving 0.60 g of nicotine (a nonelectrolyte) in water to make 12 mL of solution. The osmotic pressure of the solution is 7.55 atm at 25 °C. The molecular weight of nicotine is g/mol.
A) 43
B) 0.60
C) 160
D) 28
E) 50

Question

Quizplus · Accepted Answer

First, we need to find the molarity of the solution using the formula:

Molarity (M) = moles of solute / volume of solution in liters

We know the volume is 12 mL, which is 0.012 L. We can find the moles of solute using the formula:

moles of solute = mass of solute / molar mass

The mass of nicotine is 0.60 g. To find the molar mass, we can look it up on a periodic table or calculate it using the atomic weights of the elements in nicotine:

molar mass = (10 x 14.01 g/mol) + (14 x 1.01 g/mol) + (2 x 12.01 g/mol) + 2.00 g/mol = 160.33 g/mol

Now we can calculate the moles of nicotine:

moles of solute = 0.60 g / 160.33 g/mol = 0.003739 mol

Finally, we can calculate the osmotic pressure using the formula:

π = MRT

where π is osmotic pressure, M is molarity, R is the gas constant (0.082 L·atm/mol·K), and T is temperature in Kelvin (25°C + 273.15 = 298.15 K).

Solving for M:

M = π / RT = 7.55 atm / (0.082 L·atm/mol·K x 298.15 K) = 0.295 M

So the molarity of the nicotine solution is 0.295 M, and the molar mass of nicotine is 160 g/mol, which is choice C.

A Solution Is Prepared by Dissolving 0