Question

If \[6.055 \times {10^{ - 2}}kg\] of washing soda $(N{a_2}C{O_3}.10{H_2}O)$ is dissolved in water to obtain 1L of a solution of density $1077.2kg/{m^3}$.Calculate the molarity, molality and mole fraction of $N{a_2}C{O_3}$ in the solution.

Answer 1

Hint: In this question, we have to calculate the molarity, molality and mole fraction of sodium carbonate. As we know molality can be defined in terms of kilograms of solution and whereas the molarity is defined in terms of the value of solution in litres with respect to the moles of solute. The formula used to find molality is represented as follows:
m $ = \dfrac{{moles\;of\;solute}}{{kilograms \;solvent}}$ (m represents molality)

Complete step by step answer:
Let us first calculate the number of moles of solute
Given Mass of washing soda \[ = 6.055 \times {10^{ - 2}}kg = 60.55g\]
Molar mass of $(N{a_2}C{O_3}.10{H_2}O)$ $ = 286g$ which correspond to 1 mole of $(N{a_2}C{O_3}.10{H_2}O)$
Hence, number of moles present in the given mass of 60.55 g $ = \dfrac{{60.55}}{{286}} = 0.2117moles$
Now, we will calculate the molarity of solution i.e. M $ = \dfrac{{moles\;solute}}{{litres\;solution}}$
Therefore, $M = \dfrac{{0.2117}}{1} = 0.2117mol/L$ (Volume is 1 litre)
Hence, the molarity of solution is 0.2117 M
Now, to find molality we will use the above mentioned formula, i.e.
m $ = \dfrac{{moles\;of\;solute}}{{kilograms \;solvent}}$
 $m = \dfrac{{0.2117}}{1} = 0.2117moles/kg$ (1L water $ = $ 1 kg water)
Hence molality of the solution is 0.2117 moles /kg
Thus, to find the mole fraction, the formula used will be n $ = \dfrac{{number\; of \;molecules}}{{total\; number\;of\;molecules}}$
Or we can write it as: Mole fraction $n = \dfrac{{{n_{solute}}}}{{({n_{solute}} + {n_{solvent}})}}$
${n_{solvent}} = 1kg = 1000gofwater$
We know that molar mass of water $ = $ 18 g
Therefore, moles of solvent $ = \dfrac{{1 \times {{10}^3}}}{{18}} = 0.0555 \times {10^3}moles$
Therefore moles of solvent $ = $ 55.56 moles
Therefore, Mole fraction $n = \dfrac{{0.2117}}{{0.2117 + 55.56}} = 0.003975$
Therefore, mole fraction of the given solution is 0.003975.
In the last we can conclude that the values obtained are as follows:
Molarity $ = $ 0.2117M or 0.2117 moles/L
Molality $ = $0.2117 moles/kg
Mole fraction $ = $0.003975

Note: We have come across three different terms in the question, and the terms are different from each other and can be defined as follows:
Molarity $(M)$ is defined as the ratio of number of moles of solute present in a given volume of solution $ = $ Moles of solute/Volume of solution.
Molality $(m)$ is defined as the ratio of moles of solute to weight of solvent in kg $ = $ Moles of solute/Weight of solvent.
Mole fraction $(n)$ is defined as the ratio of number of mole of solute to number of moles of all solute plus number of moles of solvent $ = $ Number of mole of solute/Total moles of all solute present in solution.