Question

74) 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:

molar mass of nicotine will be 162.16g/mol

Explanation:

The mass of nicotine taken = 0.60g

The volume of solution = 12mL

the osmotic pressure of solution = 7.55 atm

Temperature in kelvin =298.15K (25+ 273.15)

The formula which relates osmotic pressure and concentration (moles per L) is:

π = MRT

Where

π = osmotic pressure (unit atm) = 7.55 atm

M = molarity (mol /L)

T= temperature = (K) = 298.15 K

R = gas constant = 0.0821 L atm /mol K

Putting values

$7.55=MX0.0821X298.15$

Therefore

$M=\frac{7.55}{0.0821X298.15}=0.308\frac{mol}{L}$

Molarity is moles of solute dissolve per litre of solution

The volume of solution in litre = 0.012 L

$molarity=\frac{moles}{V}$

$moles=molarityXvolume=0.308X0.012=0.0037mol$

we know that

$moles=\frac{mass}{ymolarmass}$

molar mass = $\frac{mass}{moles}=\frac{0.60}{0.0037}=162.16\frac{g}{mol}$