MHT CET · Chemistry · Some Basic Concepts of Chemistry
The percentage (by weight) of sodium hydroxide in a \(1.25\) molal \(\mathrm{NaOH}\) solution is
- A \(4.76 \%\)
- B \(1.25 \%\)
- C \(5 \%\)
- D \(40 \%\)
Answer & Solution
Correct Answer
(A) \(4.76 \%\)
Step-by-step Solution
Detailed explanation
\(1.25\) molal NaOH solution means, \(1.25\) moles of \(\mathrm{NaOH}\) are present in \(1000 \mathrm{~g}\) of water \(\therefore\) Weight of \(\mathrm{NaOH}=1.25 \times 40=50 \mathrm{~g}\)
Weight of solution \(=1000+50=1050 \mathrm{~g}\)
\(\%\) (by weight) of \(\mathrm{NaOH}\)
\(
\begin{array}{l}
=\frac{\text { wt. of solute }}{\text { wt. of solution }} \times 100 \\
=\frac{50}{1050} \times 100=4.76 \%
\end{array}
\)
Weight of solution \(=1000+50=1050 \mathrm{~g}\)
\(\%\) (by weight) of \(\mathrm{NaOH}\)
\(
\begin{array}{l}
=\frac{\text { wt. of solute }}{\text { wt. of solution }} \times 100 \\
=\frac{50}{1050} \times 100=4.76 \%
\end{array}
\)
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