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\(63 \mathrm{~g}\) of a compound (Mol. Wt. \(=126)\) was dissolved in \(500 \mathrm{~g}\) distilled water. The density of the resultant solution as \(1.126 \mathrm{~g} / \mathrm{ml}\). The molarity of the solution is
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The correct answer is:
\(1.0 \mathrm{~M}\)
Hint : Mass of compound (solute) \(=63 \mathrm{~g}\)
Mole of compound \(=\frac{63}{126}=\frac{1}{2}\) mole
\(\begin{aligned}
\text{Mass of solution}& =\text { Mass of solute }+ \text { Mass of solvent } \\
& =63+500 \\
& =563 \mathrm{~g}
\end{aligned}\)
Volume of solution \(=\frac{\text { Mass }}{\text { Density }}=\frac{563}{1.126} \mathrm{ml}\)
Molarity \(=\frac{\text { mole of compound }}{\text { volume of solution }(\text { inL })}\)
\(\begin{aligned} & =\frac{1 / 2 \times 1000}{563 / 1.126} \\ & =\frac{1.126 \times 1000}{2 \times 563}=1\end{aligned}\)
Mole of compound \(=\frac{63}{126}=\frac{1}{2}\) mole
\(\begin{aligned}
\text{Mass of solution}& =\text { Mass of solute }+ \text { Mass of solvent } \\
& =63+500 \\
& =563 \mathrm{~g}
\end{aligned}\)
Volume of solution \(=\frac{\text { Mass }}{\text { Density }}=\frac{563}{1.126} \mathrm{ml}\)
Molarity \(=\frac{\text { mole of compound }}{\text { volume of solution }(\text { inL })}\)
\(\begin{aligned} & =\frac{1 / 2 \times 1000}{563 / 1.126} \\ & =\frac{1.126 \times 1000}{2 \times 563}=1\end{aligned}\)
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