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Question: Mole fraction of \( {C_3}{H_5}{\left( {OH} \right)_3} \) in a solution of 36 g of water and 46 g of ...

Mole fraction of C3H5(OH)3{C_3}{H_5}{\left( {OH} \right)_3} in a solution of 36 g of water and 46 g of glycerine is
(A) 0.460.46
(B) 0.360.36
(C) 0.200.20
(D) 0.400.40

Explanation

Solution

Mole fraction of a substance in a mixture is the quantity or amount of the substance in the total mixture. To solve this sum, firstly we will find the moles of the solvent used. Then we will find the moles of the solute used in the process. Using the formula of mole fraction, we will finally calculate the mole fraction.

Formula used:
The formula here used to find number of moles is as follows,
Number of moles = Given weightMolecular weight\dfrac{{{\text{Given weight}}}}{{{\text{Molecular weight}}}}
The formula here used for mole fraction is as follows,
Mole fraction of a solute = Moles of soluteMoles of solute + Moles of solvent\dfrac{{{\text{Moles of solute}}}}{{{\text{Moles of solute + Moles of solvent}}}} .

Complete Step by step solution:
The formula of glycerine is C3H5(OH)3{C_3}{H_5}{\left( {OH} \right)_3} .
The given weight of water is 36 g.
Molecular weight of water is 18 g.
Moles of water = Given weight of waterMolecular weight of water=36g18g=2 moles\dfrac{{{\text{Given weight of water}}}}{{{\text{Molecular weight of water}}}} = \dfrac{{36{\text{g}}}}{{18{\text{g}}}} = 2{\text{ moles}}
The given weight of glycerine is 46 g.
Molecular weight of glycerine 92 g.
Moles of glycerine = Given weight of glycerineMolecular weight of glycerine=46g92g=0.5 moles\dfrac{{{\text{Given weight of glycerine}}}}{{{\text{Molecular weight of glycerine}}}} = \dfrac{{46{\text{g}}}}{{92{\text{g}}}} = 0.5{\text{ moles}}
Solute: It is the minor component in a solution which is dissolved in solvent.
Solvent: It is the major component in a solution in which solute is dissolved to form a solution.
Here the moles of glycerine are 0.50.5 and moles of water are 2.
Hence water is the solvent and glycerine is the solute of the solution.
Mole fraction of solute = Moles of soluteMoles of solute + Moles of solvent\dfrac{{{\text{Moles of solute}}}}{{{\text{Moles of solute + Moles of solvent}}}}
Mole fraction of glycerine = Moles of glycerineMoles of glycerine + Moles of water=0.50.5+2=0.2\dfrac{{{\text{Moles of glycerine}}}}{{{\text{Moles of glycerine + Moles of water}}}} = \dfrac{{0.5}}{{0.5 + 2}} = 0.2
Hence the mole fraction of C3H5(OH)3{C_3}{H_5}{\left( {OH} \right)_3} in the solution is 0.200.20 .
Therefore, the correct option is C.

Note:
Mole fraction is basically the ratio of moles. Hence it is unitless.
The mole fraction of the total solution is 1 and the mole fraction of solute or solvent separately is always less than 1.