Question

Consider a compound “A” which dissolved in water in equimolar ratio and formed a solution of average molar mass of $200g/mol$. If the same compound “A” forms a $19.1\%\left(w/v\right)$ solution with another solvent 'X' then the molarity of solution 'A' with 'X' would be equal to:

• A. $0.005M$
• B. $5M$
• C. $0.05M$
• D. $0.5M$

Answer 1

The correct option is D $0.5M$

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 $0.5$
$\text{Average molar mass}=(\text{mole fraction of solute}\times \text{molar mass of solute})+(\text{mole fraction of solvent}\times \text{molar mass of solvent})$
$200=0.5\times M_A+0.5\times 18$
On solving,
$M_A=382g/mol$
Case 2:
$19.1\%\left(w/v\right)$ means $19.1g$ of solute is present in $100mL$ of solution.
$\text{Moles of A in 100 mL of solution}=\frac{19.1}{382}=0.05mol$

$\text{Molarity of solution}=\frac{\text{moles of solute}}{\text{volume of solution in mL}}\times 1000$

$\text{Molarity}=\frac{0.05}{100}\times 1000=0.5M$