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 $\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Liter}\text{of}\text{soln}\text{.}}$

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

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