Question

What will be the molarity of a solution, which contains $5.85\, g$ of $NaCl_(s)\, per\, 500\, mL$?

• A. $ 4$ mol $L^{-1}$
• B. $ 20$ mol $L^{-1}$
• C. $ 0.2$ mol $L^{-1}$
• D. $ 2$ mol $L^{-1}$

Answer 1

Answer: (C)

Molarity $(M)= \frac{\text{No. of moles of solute }}{\text{Volume of solution (in $L$) }}$
or, Molarity $= \frac{W_B \times 1000}{M_B {\text{ Volume of solution (in $mL$)}}} $
$ =\frac{5.85 \times1000}{58.5\times500} $ ($\therefore $ Molar mass of $NaCl = 58.5\, g\, mol^{-1}$)
$= 0.2\, mol\, L^{-1}$