Question

 Determine the concentration of a solution (in % by mass, % by volume, % by mass

volume, molarity, molality and ppm) containing 5 grams of NaCI and 50 ml of water.(p of NaCl = 2.1 g/ml,p of water = 1g/ml).What is the mole fraction of NaCI in the solution?

Answer 1

Ans:-

Given $5$ grams of $NaCI$ and $50 ml$ of water

Moles of $NaCl=\dfrac{5}{58.4}=0.0856mol$

Concentration of Solution in molarity$=\dfrac{Moles of Sollution}{volume of solution}=\dfrac{0.0856}{50}\times1000=1.7123M$

Concentration of Solution in molality$=\dfrac{moles of solute}{Mass of Solvent}=\dfrac{0.0856}{50}\times1000=1.7123m$

We know that density of Water is $1g/ml$.

So, mass of Water present in solution $=$ density$\times$ volume$=1\times50=50g$

Hence moles of water$=\dfrac{50}{18}=2.77mol$

Mole fraction of $NaCl$ in solution$= \dfrac{0.0856}{0.0856+2.77}=0.03$