Answer to Question #172529 in General Chemistry for Blank 0822

Question #172529

40.0 grams of Calcium Carbonate compound (CaCO3) with molar mass of 100g/mole, in order to make an aqueous solution that has a mole fraction of solute of 0.200? How much H2O is needed to achieved the desired result? 


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Expert's answer
2021-03-22T06:18:02-0400

Q172529

40.0 grams of Calcium Carbonate compound (CaCO3) with a molar mass of 100g/mole, in order to make an aqueous solution that has a mole fraction of solute of 0.200? How much H2O is needed to achieve the desired result? 


Solution:


We will first convert 40.0 g of CaCO3 to moles by using the molar mass 100 g/mole.

Here CaCO3 is the solute and H2O is the solvent.

By using the moles of CaCO3 and the given mole fraction of solute (CaCO3) we can find the total moles in the solution.

Next from the total moles and moles of CaCO3, we can find the moles of solvent (water ).

Lastly, we can convert moles of H2O to grams of H2O.


Step 1: Convert 40.0 grams of CaCO3 to moles by using the molar mass 100 g/mole.



moles of CaCO3=40.0g of CaCO31 mol CaCO3100 g CaCO3moles \space of \space CaCO_3 = 40.0g \space of\space CaCO_3 * \frac{ 1\space mol \space CaCO_3 }{100 \space g \space CaCO_3 }

=0.40 mol of CaCO3= 0.40\space mol\space of \space CaCO_3

Step 2: To find total moles in solution by using 0.40 mol CaCO3 and

mole fraction of solute CaCO3 = 0.200.




mole fraction of CaCO3=moles of CaCO3total moles in solutionmole \space fraction \space of \space CaCO_3 = \frac{moles \space of \space CaCO_3 }{ total \space moles \space in \space solution }

Arranging this equation for Total moles, we have



total moles in solution=moles of CaCO3mole fraction of CaCO3total \space moles \space in \space solution = \frac{moles \space of \space CaCO_3 }{mole \space fraction \space of \space CaCO_3 }



=0.40 mol0.20=2.00moles= \frac{0.40\space mol }{0.20 } = 2.00 moles

Hence total moles of solute and solvent in the solution is 2.00 moles.



Step 3: To find moles of H2O (solvent)


moles of solution = moles of solute ( CaCO3 ) + moles of solvent ( H2O).


Substitute moles of solution = 2.00 moles, and moles of solute = 0.40 moles, we have


2.00 mol = 0.40 mol + moles of solvent ( H2O).


moles of solvent ( H2O) = 2.00 mol - 0.40 mol = 1.60 moles.


Hence moles of solvent ( water ) in the solution is 1.60 moles.


Step 4: To convert moles of H2O to grams by using its molar mass.


molar mass of H2O = 2 * atomic mass of H + 1 * atomic mass of O

= 2 * 1.00794 g/mol + 1 * 15.999 g/mol

= 2.01588 g/mol + 15.999 g/mol

= 18.015 g/mol.



grams of H2O=1.60 mol H2O18.015 g of H2O1 mol of H2Ograms\space of \space H_2O = 1.60\space mol \space H_2O * \frac{18.015\space g \space of\space H_2O }{1\space mol\space of\space H_2O }


=28.824 grams of H2O= 28.824\space grams \space of\space H_2O

in the given question we are given all the quantities in 3 significant figure, so our final answer must also be in 3 significant figure.


So the mass of water required to achieve the desired result is 28.8 grams.


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