Solveeit Logo

Question

Question: Given, \(\dfrac{w}{v}\% \) of urea solution is \(6.3\% \) its density is \(1.05\;gm\;m{l^{ - 1}}\). ...

Given, wv%\dfrac{w}{v}\% of urea solution is 6.3%6.3\% its density is 1.05  gm  ml11.05\;gm\;m{l^{ - 1}}. Mole fraction of urea is nearly
A.0.0180.018
B.0.0190.019
C.0.0240.024
D.0.0300.030

Explanation

Solution

The quantity of molecules of a specific component in a mixture divided by the total quantity of moles in the given mixture is represented by mole fraction. It is one process which expresses the concentration of a solution.
Mole Fraction formula:
mole  fraction  ofsolute=moles  of  solutemole  of  solute+mole  ofsolventmole\;fraction\;of\,solute = \dfrac{{moles\;of\;solute}}{{mole\;of\;solute + mole\;of\,solvent}}
=nAnA+nB= \dfrac{{{n_A}}}{{{n_A} + {n_B}}}

Complete answer:
In chemistry, the amount of component ni{n_i} divided by the total amount of all component of the mixture ntot{n_{tot}} is referred to mole fraction or molar fraction xi{x_i} or Xi{X_i}.
xi=nintot{x_i} = \dfrac{{{n_i}}}{{{n_{tot}}}}
The summation of all the mole fraction is equal to 11:
i=1Nni=ntot\sum\limits_{i = 1}^N {{n_i}} = {n_{tot}} , i=1Nxi=1\sum\limits_{i = 1}^N {{x_i}} = 1
Another term used for mole fraction is amount fraction. It is similar to the number fraction which can be explained as the number of molecules of a constituent Ni{N_i} divided by the total number of all molecules Ntot{N_{tot}}. The composition of a mixture with a dimensionless quantity; mass fraction and volume fraction are others is expressed by one method which is mole fraction.
Mass fraction wi{w_i} is the product of the ratio of mass of one component to the total mass of the all component and mole fraction.
It can be found by using the formula,
wi=xiMiM=xIMijxjMj{w_i} = {x_i}\dfrac{{{M_i}}}{{\overline M }} = {x_I}\dfrac{{{M_i}}}{{\sum\nolimits_j {{x_j}} {M_j}}}
Here, Mi{M_i} is the molar mass of the component ii and M\overline M is the average molar mass of the mixture.
According to the question:
wv%=6.3%\dfrac{w}{v}\% = 6.3\%
This means that 6.3  g6.3\;g urea in 100  ml100\;ml solution
since,
massofsolution=density×volumemass\,of\,solution = density \times volume
=1.05×100= 1.05 \times 100
=105  g= 105\;g
Thus,
massofwater(solvent)=1056.3mass\,of\,water\left( {solvent} \right) = 105 - 6.3
=98.7  g= 98.7\;g
moles  of  urea=massmolar  massmoles\;of\;urea = \dfrac{{mass}}{{molar\;mass}}
=6.360= \dfrac{{6.3}}{{60}}
=0.105= 0.105
molesofwater=98.718moles\,\,of\,water = \dfrac{{98.7}}{{18}}
=5.483= 5.483
Xurea=nureanurea+nwater{X_{urea}} = \dfrac{{{n_{urea}}}}{{{n_{urea}} + {n_{water}}}}
=0.1050.105+5.483= \dfrac{{0.105}}{{0.105 + 5.483}}
Xurea=0.0187{X_{urea}} = 0.0187
Thus, this is the required answer.

Note:
Please remember that mole fraction stands in for a fraction of molecules. As different molecules consist of different masses and there is the difference in mole fraction from the mass fraction.