The concentration of a solution obtained by dissolving \(0.53 g\) of pure anhydrous \(Na _{2} CO _{3}\) in water to make \(250 cm ^{3}\) of solution is ____________
A. \(2.0 \times 10^{-5} mol dm ^{-3}\) B. \(2.1 gdm ^{-3}\) C. \(2.0 \times 10^{-}\) D. \(5.0 \times 10^{-5} mol dm^{-3 }\)
Correct Answer: C
Explanation
\(0.53 g\) of \(Na _{2} CO _{3}\) is dissolved in \(250 cm ^{3}\) then \(x g\) would dissolve in \(1000 cm ^{3}\left(1 dm ^{3}\right)\) i.e. \begin{aligned} &\begin{array}{l}53 g \rightarrow 250 cm ^{3} \\ x g \rightarrow 1000 cm ^{3} \end{array} \\ &x=\frac{0.53 \times 1000}{250}=2.12 g \end{aligned} that is \(2.12 g / dm ^{3}\), \begin{aligned} & MMNa _{2} CO _{3}=106 g / mol \\ &(\text { Molar conc. })=\frac{\text { Mass conc. }}{\text { Molar Mass }} \\ &=\frac{2.12 g / dm ^{3}}{106 g / mol }=0.02 M \\ &=2.0 \times 10^{-2} mol / dm ^{3} \end{aligned}