Answer to Question #249067 in General Chemistry for Seven

Question

A solution of calcium carbonate, CaCO3, in which the mole fraction of the solute is 0.19. What mass of CaCO3, in grams, should you combine with 155.0g of water? Molar mass of CaCO3 is 100.09 g/mol.

Answer 1

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

$$moles \space of \space CaCO_3 = 40.0g \space of\space CaCO_3 * \frac{ 1\space mol \space CaCO_3 }{100 \space g \space CaCO_3 }$$ $$= 0.40\space mol\space of \space CaCO_3$$

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

mole fraction of solute CaCO₃ = 0.200.

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

Arranging this equation for Total moles, we have

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

$$= \frac{0.40\space mol }{0.20 } = 2.00 moles$$

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.

$$grams\space of \space H_2O = 1.60\space mol \space H_2O * \frac{18.015\space g \space of\space H_2O }{1\space mol\space of\space H_2O }$$ $$= 28.824\space grams \space of\space H_2O$$

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.