Question

Which of the following gives the molarity of a 17.0% by mass solution of sodium acetate, $\mathrm{CH}_3 \mathrm{COONa}(\mathrm{FM}=82.0 \mathrm{amu})$ in water? in water? Given the density is $\text{1} \text{.09} \, \text{g/mol}$ .

• A. $2.26\times 10^{- 6}\text{M}$
• B. 0.207 M
• C. 2.07 M
• D. 2.26 M

Answer 1

Answer: (D)

$\text{17} \text{.0\%}$ solution of $\text{CH}_{\text{3}} \text{COONa}$ means that $\text{100} \, \text{g}$ solution contain $\text{17} \, \text{g}$ of $\text{CH}_{\text{3}} \text{COONa}$ .

Moles of $\text{CH}_{\text{3}} \text{COONa} \, \text{=} \, \frac{\text{17} \, \text{g}}{\text{82} \text{.0} \, \text{g} \, \text{mol}^{- \text{1}}} \, \text{=} \, \frac{\text{17}}{\text{82} \text{.0}} \, \text{mol}$

Volume of $\text{100} \, \text{g}$ solution $=\frac{\text{Mass of solution}}{\text{Density of solution}}$

$\text{=} \, \frac{\text{100} \, \text{g}}{\text{1} \text{.09} \, \text{g} \, \text{mol}^{- \text{1}}} \, \text{=} \, \frac{\text{100}}{\text{1} \text{.09}} \text{mL}$

So molarity of solution $\text{=} \, \frac{\text{Moles} \, \text{of} \, \text{solute}}{\text{Liter} \, \text{of} \, \text{soln} \text{.}}$

$=\frac{\frac{17}{82.0} mol }{\left(\frac{100}{1.09}\right) \times 10^{-3} L }$

$\text{=} \, \text{2} \text{.259} \, \text{M} \, \text{=} \, \text{2} \text{.26} \, \text{M}$