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ChemEasyAdv 2019 • Paper 2

Molarity Calculation of Aqueous Urea Solution from Mole Fraction and Density

In an aqueous solution containing 900 g900\text{ g} of water, the mole fraction of urea is found to be 0.050.05. If the measured density of this solution is 1.2 g cm−31.2\text{ g cm}^{-3}, calculate the molarity of the urea solution. (Given data: Molar masses of urea and water are 60 g mol−160\text{ g mol}^{-1} and 18 g mol−118\text{ g mol}^{-1}, respectively)

Official Numerical Answer2.98 to 2.99

Topics & Concepts

SolutionsConcentration TermsMole Concept

Step-by-Step Solution

To find the molarity of the urea solution, we follow these steps:

Step 1: Calculate the number of moles of water

Given:

  • Mass of water (wwaterw_{\text{water}}) = 900 g900\text{ g}
  • Molar mass of water (MwaterM_{\text{water}}) = 18 g mol−118\text{ g mol}^{-1}

Moles of water (nwater)=wwaterMwater=900 g18 g mol−1=50 mol\text{Moles of water } (n_{\text{water}}) = \frac{w_{\text{water}}}{M_{\text{water}}} = \frac{900\text{ g}}{18\text{ g mol}^{-1}} = 50\text{ mol}

Step 2: Calculate the number of moles of urea

Given that the mole fraction of urea (XureaX_{\text{urea}}) is 0.050.05:

Xurea=nureanurea+nwater=0.05X_{\text{urea}} = \frac{n_{\text{urea}}}{n_{\text{urea}} + n_{\text{water}}} = 0.05

nureanurea+50=0.05\frac{n_{\text{urea}}}{n_{\text{urea}} + 50} = 0.05

nurea=0.05 nurea+2.5n_{\text{urea}} = 0.05\,n_{\text{urea}} + 2.5

0.95 nurea=2.5  ⟹  nurea=2.50.95=5019 mol0.95\,n_{\text{urea}} = 2.5 \implies n_{\text{urea}} = \frac{2.5}{0.95} = \frac{50}{19}\text{ mol}

Step 3: Calculate the total mass of the solution

Given:

  • Molar mass of urea (MureaM_{\text{urea}}) = 60 g mol−160\text{ g mol}^{-1}

Mass of urea (wurea)=nurea×Murea=5019 mol×60 g mol−1=300019 g\text{Mass of urea } (w_{\text{urea}}) = n_{\text{urea}} \times M_{\text{urea}} = \frac{50}{19}\text{ mol} \times 60\text{ g mol}^{-1} = \frac{3000}{19}\text{ g}

The total mass of the solution (msolm_{\text{sol}}) is:

msol=wwater+wurea=900+300019=17100+300019=2010019 gm_{\text{sol}} = w_{\text{water}} + w_{\text{urea}} = 900 + \frac{3000}{19} = \frac{17100 + 3000}{19} = \frac{20100}{19}\text{ g}

Step 4: Calculate the volume of the solution

Given:

  • Density of the solution (dd) = 1.2 g cm−3=1.2 g mL−11.2\text{ g cm}^{-3} = 1.2\text{ g mL}^{-1}

Volume of the solution (Vsol)=msold=2010019 g1.2 g mL−1=2010019×1.2 mL=2010022.8 mL\text{Volume of the solution } (V_{\text{sol}}) = \frac{m_{\text{sol}}}{d} = \frac{\frac{20100}{19}\text{ g}}{1.2\text{ g mL}^{-1}} = \frac{20100}{19 \times 1.2}\text{ mL} = \frac{20100}{22.8}\text{ mL}

Converting the volume into liters:

Vsol=2010019×1200 L=201228 L=6776 LV_{\text{sol}} = \frac{20100}{19 \times 1200}\text{ L} = \frac{201}{228}\text{ L} = \frac{67}{76}\text{ L}

Step 5: Calculate the molarity of the urea solution

Molarity (MM) is defined as the number of moles of solute per liter of solution:

M=nureaVsol (in L)=50196776=5019×7667M = \frac{n_{\text{urea}}}{V_{\text{sol}}\text{ (in L)}} = \frac{\frac{50}{19}}{\frac{67}{76}} = \frac{50}{19} \times \frac{76}{67}

Since 7619=4\frac{76}{19} = 4:

M=50×467=20067 mol L−1≈2.985 MM = \frac{50 \times 4}{67} = \frac{200}{67}\text{ mol L}^{-1} \approx 2.985\text{ M}

Rounding off to two decimal places, the molarity of the urea solution is 2.992.99 (or 2.982.98).

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