Question

The mole fraction of urea in aqueous urea containing $900g$ of water is $0.05$. If the density of the solution in $1.2g/c{{m}^{3}}$, the molarity of the urea solution is _______.
Given: Molar mass of urea = $60gmo{{l}^{-1}}$

Answer 1

Hint: Consider the formula for molarity and how the molarity depends on the number of moles of solute and the volume of the solution. Find the volume of the solution using the mass and the density of the solution. Try to find the value of the number of moles f urea using the information given about the mole fraction.

Complete step by step solution:
We know that the general formula for the molarity of a solution is dependent on the volume of the solution and the number of moles of the solute. Both of these values are not given in the question and we will have to find them from the given information. The formula for molarity is:
\[\begin{align}
& \text{molarity = }\dfrac{\text{number of moles of solute}}{\text{volume of solution}} \\
& M=\dfrac{{{n}_{u}}}{{{V}_{s}}} \\
\end{align}\]
First, we will find the number of moles of urea using the given information about the mole fraction. We know that the mole fraction of the solute is the number of moles of the solute divided by the total number of moles of solute and solvent. Let ${{n}_{u}}$ be the number of moles of urea, ${{n}_{w}}$ be the number of moles of water, and ${{\chi }_{u}}$ be the mole fraction of urea. The formula will be:
\[{{\chi }_{u}}=\dfrac{{{n}_{u}}}{{{n}_{u}}+{{n}_{w}}}\]
The number of moles of water will be the given mass of water $(900g)$ divided by the molar mass of water $(18gmo{{l}^{-1}})$. Now, we will put the values that we have in the formula and solve for ${{n}_{u}}$.
\[\begin{align}
& 0.05=\dfrac{{{n}_{u}}}{{{n}_{u}}+\dfrac{900}{18}} \\
&\Rightarrow 0.05({{n}_{u}}+50)={{n}_{u}} \\
&\Rightarrow 0.05{{n}_{u}}+2.5={{n}_{u}} \\
& \Rightarrow 0.95{{n}_{u}}=2.5 \\
& \Rightarrow {{n}_{u}}=2.631mol \\
\end{align}\]
We now have to find the volume of the solution. The formula that relates density mass and volume can be used here, we have the density and to find the mass we need to add the mass of the solute to the mass of the solvent that is given. We can find the mass of the solute from the number of moles of solute.
\[\begin{align}
& {{n}_{u}}=\dfrac{\text{weight of urea}}{\text{molar mass of urea}} \\
& \Rightarrow \text{weight of urea}=2.631\times 60 \\
& \Rightarrow \text{weight of urea}=157.86g \\
\end{align}\]
Adding this to the weight of the solvent, we will get the weight of the solution. We will find the volume using the formula for density.
\[\begin{align}
& density=\dfrac{mass}{volume} \\
& \Rightarrow 1.2g/c{{m}^{3}}=\dfrac{900+157.86g}{{{V}_{s}}} \\
& \Rightarrow {{V}_{s}}=\dfrac{1057.86}{1.2}c{{m}^{3}} \\
& \Rightarrow {{V}_{s}}=881.55c{{m}^{3}} \\
\end{align}\]
We will now put these values in the formula for molarity and obtain its value.
\[\begin{align}
& molarity=\dfrac{moles}{volume}\times 1000 \\
& \Rightarrow M=\dfrac{2.631}{881.55}\times 1000 \\
& \Rightarrow M=2.98M \\
\end{align}\]

Hence, the molarity of the solution is approximately 3M.

Note: To calculate the molarity, we are multiplying the traditional formula by 1000 since the value of the volume is required to be in liters. We have obtained the value in cubic centimeters, so to compensate for that, we are multiplying the formula by a factor of 1000.