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JEE Main 202430 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2024 · 30 January · Evening Shift · Chemistry · Solutions

If a substance ' A ' dissolves in solution of a mixture of ' B '...

If a substance ′A′' A ' dissolves in solution of a mixture of ' BB' and ' CC ' with their respective number of moles as nA,nB{n}_{A}, {n}_{B} and nC{n}_{C}, mole fraction of CC in the solution is:

Options

  1. AnCnA×nB×nC\dfrac{{n}_{C}}{{n}_{A} \times {n}_{B} \times {n}_{C}}
  2. BnCnA+nB+nC\dfrac{{n}_{C}}{{n}_{A} + {n}_{B} + {n}_{C}}
  3. CnCnA−nB−nC\dfrac{{n}_{C}}{{n}_{A} - {n}_{B} - {n}_{C}}
  4. DnBnA+nB\dfrac{{n}_{B}}{{n}_{A} + {n}_{B}}

Correct answer

B. nCnA+nB+nC\dfrac{{n}_{C}}{{n}_{A} + {n}_{B} + {n}_{C}}

Step-by-step solution

Mole fraction is defined as the amount (number of moles or molecules) of a constituent or component divided by the total amount of all the constituent or component present. 

Mole fraction=NumberofmolesofcompnentTotalnumberofmolesofallthecomponents= \dfrac{Numberofmolesofcompnent}{Totalnumberofmolesofallthecomponents}
Suppose we have a mixture with 33 components A, B and C.

Mole fraction of CC=nCnA+nB+nC= \dfrac{{n}_{C}}{{n}_{A} + {n}_{B} + {n}_{C}}

Hence, the answer is option B.

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