Question

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.

Answer 1

Answer:

Molecular weight of nicotine is 162 g/mol

Explanation:

Osmotic pressure ($\pi$) for nonelectrolyte is expressed as-

                                    $\pi =CRT$

where C is molarity of solution, R is gas constant and T is temperature in Kelvin scale.

If molecular weight of nicotine is W g/mol then number of moles of nicotine present in solution is equal to $\frac{0.60}{W}$ moles

So, molarity (C) of solution = $\frac{0.60\times 1000}{W\times 12}M$

R is equal to 0.082 L.atm/(mol.K), T is 298 K and $\pi$) is 7.55 atm

So, $7.55=(\frac{0.60\times 1000}{W\times 12})\times (0.082)\times (298)$

or, W = 162

So molecular weight of nicotine is 162 g/mol