Playing with Concentration 2

1000g solution-->3mol solute=138g.
Total mass=1138 g.
Mass/density=volume.
Vol=1138/1.6 mL.
M=mol/vol (L)=3×1.6×1000/1138=4.22

Consider the $\text{3 molal}$ alcohol solution with $\text{1 kg}$ of water.

The mass of $\ce{C2H6O}$ in the solution $m_{\ce{C2H6O}} = 3 \times \text{molar mass of } \ce{C2H6O}$ $= 3(12\times 2 + 1\times 4 + 16)$ $= \text{138 g}$

Therefore, the mass of the solution $m_{sol} = m_{\ce{C2H6O}} + m_{\ce{H2O}} = 138 + 1000 = \text{1138 g}$

The volume of the solution $V_{sol} = \dfrac{1138}{1.6 \times 1000} \text{ L}$ .

Therefore, molarity of the solution $M_{sol} = \dfrac {\text{3 mol}}{V_{sol}} = \dfrac{3 \times 1600}{1138} = \boxed{\text{4.218 mol/L}}$

I really do like this problem.

3 mol / (1000 g [solvent])

3 mol / (1000 g + (46 g) * 3 mol) = 3 mol / (1138 g)

2.6362 mol / g [solution]

2.6362 mol / g * (1.6 g / L) = 4.22 mol / L