Question

Consider a hypothetical case where a compound “A” is dissolved in water in equimolar ratio and formed a solution of average molar mass 100 g/mol. If same compound “A” formed 18.2% (w/v) solution with another solvent X then molarity of solution (A with X) would be equal to:

• A. 0.001 M
• B. 1 M
• C. 0.01 M
• D. 0.1 M

Answer 1

The correct option is B 1 M

Let the molar mass of compound A be $M_A$ g
Case 1:
Both solute A and solvent are present in equimolar ratio. So, mole fraction of each will be 0.5
$Averagemolarmass=molefractionofsolute\times molarmassofsolute)+(molefractionofsolvent\times molarmassofsolvent)$
$100=0.5\times M_A+0.5\times 18$
On solving, $M_A=182g/mol$
Case 2:
18.2% (w/v) means, 18.2 g of solute is present in 100 mL of solution.
Moles of A in 100 mL solution$=\frac{18.2}{182}=0.1mol$
$Molarityofsolution=\frac{molesofsolute}{volumeofsolutioninmL}\times 1000$
$Molarity=\frac{0.1}{100}\times 1000=1M$