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Question: In an experiment, \(18.0g\) of mannitol dissolves in \(100g\) of water. The vapour pressure of solut...

In an experiment, 18.0g18.0g of mannitol dissolves in 100g100g of water. The vapour pressure of solution at 20c20{}^\circ cwas 17.226mm17.226mm of mercury. Calculate the molecular mass of mannitol. The vapour pressure of water at 20c20{}^\circ c is 17.535mm17.535mm of mercury.
A.Molecular mass of mannitol = 171.02171.02
B.Molecular mass of mannitol = 181.02181.02
C.Molecular mass of mannitol =161.02161.02
D.Molecular mass of mannitol =121.02121.02

Explanation

Solution

A solution is formed, when one substance dissolves into another. A solution is a homogeneous mixture of a solvent-dissolved solution. The solute is the dissolved substance whereas the dissolving medium is the solvent. Both the solute and solvent are liquid and use the reduction of vapour pressure which is Δp1=p10p1\Delta {{p}_{1}}=p_{1}^{0}-{{p}_{1}} and calculate it by using mole fraction.

Complete answer:
Let us know some points first that liquid solutions are formed when solvent is a liquid. The solute can be gas, liquid or solid to make liquid solution. Here Mannitol and water both are liquid which means solute and solvent both are in liquid state. Now let us consider it as a binary solution (the solutions containing two components) of liquid in liquid.
As we want to find the molecular mass of mannitol we are going to use Raoult’s Law which states that the partial pressure (Vapour pressure) of each and every component of the solution is ∝(directly proportional) to its mole fraction present in solution.
So we can say that for first component
p1 x1{{p}_{1}}\propto ~{{x}_{1}}
p1=p10x1{{p}_{1}}=p_{1}^{0}{{x}_{1}}
Now for second we will do the same
p2=p20x2{{p}_{2}}=p_{2}^{0}{{x}_{2}}
ptotat=p1+p2=p10x1+p20x2\therefore {{p}_{totat}}={{p}_{1}}+{{p}_{2}}=p_{1}^{0}{{x}_{1}}+p_{2}^{0}{{x}_{2}}
Now if we want to find the reduction in the vapour pressure of solvent it is given as:
Δp1=p10p1\Delta {{p}_{1}}=p_{1}^{0}-{{p}_{1}}
To find the value we need to find value we need to first know what information we have now
Mass of solute = 18.0g18.0g
Mass of water = 100g100g
So, No. of moles of solute will be = 18Molecular weight\dfrac{18}{Molecular\text{ }weight}
No. of moles of water will be = 10018=5.55(approx)\dfrac{100}{18}=5.55(approx)
Now from equation
p1=p10x1{{p}_{1}}=p_{1}^{0}{{x}_{1}} Or p10p1=p10x1p_{1}^{0}-{{p}_{1}}=p_{1}^{0}{{x}_{1}}
By putting the values, we can say that
(17.53517.226)=17.535×x1(17.535-17.226)=17.535\times {{x}_{1}}
x1=0.30917.535{{x}_{1}}=\dfrac{0.309}{17.535}
Now The mole fraction will be =18M18M+5.55=0.0176=\dfrac{\dfrac{18}{M}}{\dfrac{18}{M}+5.55}=0.0176 ( here ‘M’ is Molar Weight)
Now by solving it we get
M=181.02M=181.02

So, Option B is the correct answer.

Note:
The properties of solution which are dependent only on the total number of particles or total concentration of particles in the solution and they are not dependent on the nature of particles like shape, size etc. are called colligative properties.
Whereas the properties which are dependent on the nature of particles are constitutional properties like electrical conductance.
There are 4 colligative properties of solution in which one of them is Relative lowering in vapour pressure(Δpp)\left( \dfrac{\Delta p}{p} \right).