02. SOLUTIONS (HM)
Explanation:
It is given that the density of the solution is \(1.110 \,g\, mL ^{-1}\).
The concentration of sodium hydroxide solution is \(3.0\) molal.
This implies \(1000 \,g\) of water has \(3\) moles of \(NaOH\).
\(3\) moles of \(NaOH =3 \times 40=120 \,g\) of \(NaOH\)
Mass of solution \(=1000+120=1120\, g\)
Volume of solution \(=\frac{\text { Mass }}{\text { Density }}=\frac{1120}{1.11}=1009\, mL\)
Molarity \(=\frac{\text { Number of moles }}{\text { Volume in L }}=\frac{3}{\frac{1009}{1000}}=2.97\, M\)
Hence, the molarity of the solution is \(2.97\, M\).