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 ____________
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Correct Answer: Option 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}
\(0.53 g\) 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}