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Stoichiometry

Q1: A very dilute solution is prepared by dissolving 'x1' mole of solute in 'x2' mole of a solvent. The mole fraction of solute is approximately equal to

A x1/x2

B x2/x1

C 1 - (x1/x2)

D 1/x2

ANS:A - x1/x2

To find the mole fraction of the solute (𝑋1X1​), we use the formula: 𝑋1=moles of solutetotal moles in solutionX1​=total moles in solutionmoles of solute​ In this case, the total moles in the solution is the sum of the moles of solute (𝑥1x1​) and the moles of solvent (𝑥2x2​). So, the mole fraction of the solute (𝑋1X1​) is: 𝑋1=𝑥1𝑥1+𝑥2X1​=x1​+x2​x1​​ Thus, the mole fraction of the solute is approximately equal to 𝑥1𝑥1+𝑥2x1​+x2​x1​​. Therefore, the correct option is 𝑥1𝑥1+𝑥2x1​+x2​x1​​.



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