question_answer 20)

If the density of methanol is \[0.793\,\,kg\,\,{{L}^{-1}}\]. What is its volume needed for making 2.5L of its 0.25M solution?

Answer:

We know, \[M=\frac{{{w}_{B}}\times 1000}{{{m}_{B}}\times V}\,\,\,\,\,\,\,\,\,\,\,\,......(i)\] Given : M = 0.25 g, V = 2.5L = 2500 mL, \[{{m}_{B}}\] (molar mass of \[C{{H}_{3}}OH\]) = 32 g \[mo{{l}^{-1}}\] Putting these values in Eqn. (i), we get \[0.25=\frac{{{w}_{B}}\times 1000}{32\times 2500}\] \[{{w}_{B}}=20g\] Volume of solute = \[\frac{Mass}{Density}=\frac{20}{0.793}\] = 25.22 mL (Here, density = \[0.793kg\,{{L}^{-1}}\] or 0.793 g \[0.793g\,m{{L}^{-1}}\])