The molarity of one molal glucose solution having density of $1.2 \mathrm{~g} / \mathrm{mL}$ is
The molarity of one molal glucose solution having density of $1.2 \mathrm{~g} / \mathrm{mL}$ is
- $0.101 \mathrm{M}$
- $1.01 \mathrm{M}$
- $2.01 \mathrm{M}$
- $0.001 \mathrm{M}$
Solution
$\mathrm{d}=1.2 \mathrm{~g} \mathrm{~mL}^{-1}, \mathrm{~m}=1 \mathrm{~mol} \mathrm{~kg}$.
Let's take $1 \mathrm{~kg}$ of the solution.
$
\begin{aligned}
& \Rightarrow \text { Volume of the solution }=\frac{\text { mass }}{\text { density }} \\
& =\frac{1000 \mathrm{~g}}{1.2 \mathrm{~g} \mathrm{~mL}^{-1}}=833.33 \mathrm{~mL} \\
& =0.833 \mathrm{~L} \\
& \Rightarrow \text { Number of moles }=1.0 \mathrm{~mol}
\end{aligned}
$
Therefore, molarity $=\frac{1.0 \mathrm{~mol}}{0.833 \mathrm{~L}}$ $=1.20 \mathrm{M} \approx 1.01 \mathrm{M}$
Asked in: AP EAMCET 2023 (15 May Shift 1)
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