Question

Calculate the mole fraction of each solute and solvent:

1. 583 g of\({{\bf{H}}_2}{\bf{S}}{{\bf{O}}_4}\)in 1.50 kg of water—the acid solution used in an automobile battery
2. 0.86 g of NaCl in 1.00 × 102 g of water—a solution of sodium chloride for intravenous injection
3. 46.85 g of codeine, \({{\bf{C}}_{{\bf{18}}}}{{\bf{H}}_{{\bf{21}}}}{\bf{N}}{{\bf{O}}_{\bf{3}}}\), in 125.5 g of ethanol, \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH}}\)
4. 25 g of I2 in 125 g of ethanol, \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH}}\).

Short answer

1. Mole Fraction of \({{\bf{H}}_2}{\bf{S}}{{\bf{O}}_4}\) and \({{\bf{H}}_{\bf{2}}}{\bf{O}}\) are 0.066 and 0.933 respectively.
2. Mole Fraction of each solute and solvent -Mole fractions of NaCl and\({{\bf{H}}_{\bf{2}}}{\bf{O}}\)are 0.003 and 0.997 respectively.
3. Mole Fraction of each solute and solvent -Mole fractions of Codeine and Ethanol are 0.054 and 0.945 respectively.
4. Mole Fraction of each solute and solvent

Mole fractions of I₂ and Ethanol are 0.035 and 0.96 respectively.

Step-by-step solution

Mole fraction

A mole fraction may be defined as the number of moles of component and the total number of moles of moles of solution.

Mole Fraction of \({{\bf{H}}_2}{\bf{S}}{{\bf{O}}_4}\) and \({{\bf{H}}_{\bf{2}}}{\bf{O}}\)

Firstly, we can find number of moles from mass and molar mass

Mass of \({{\bf{H}}_2}{\bf{S}}{{\bf{O}}_4}\)= 583g

Molar Mass of \({{\bf{H}}_2}{\bf{S}}{{\bf{O}}_4}\) = 98 g/mole

Now, number of Moles is

\({{\bf{H}}_2}{\bf{S}}{{\bf{O}}_4}\)= \(\frac{{583}}{{98}} = 5.95\)

Mass of Water = 1500g

Molar Mass of Water = 18 g/mole

Now, number of moles is

\({{\bf{H}}_{\bf{2}}}{\bf{O}}\)= \(\frac{{1500}}{{18}} = 83.33\)

Now, we can find out mole fraction of each component

\({{\bf{H}}_2}{\bf{S}}{{\bf{O}}_4}\)= \(\frac{{5.95}}{{5.95 + 89.28}} = 0.066\)

\({{\bf{H}}_{\bf{2}}}{\bf{O}}\)= \(\frac{{83.33}}{{5.95 + 83.33}} = 0.933\)

Mole Fraction of NaCl and \({{\bf{H}}_{\bf{2}}}{\bf{O}}\)

Mass of NaCl= 0.86g

Molar Mass of NaCl = 58.5 g/mole

Number of Moles of \({\bf{NaCl = }}\frac{{{\bf{0}}{\bf{.86}}}}{{{\bf{58}}{\bf{.5}}}}{\bf{ = 0}}{\bf{.0147}}\)

Mass of Water = 102g

Molar Mass of Water = 18 g/mole

Number of \({\bf{Moles = }}\frac{{{\bf{102}}}}{{{\bf{18}}}}{\bf{ = 5}}{\bf{.66}}\)

Mole Fraction of \({\bf{NaCl = }}\frac{{{\bf{0}}{\bf{.0147}}}}{{{\bf{0}}{\bf{.0147 + 5}}{\bf{.66}}}}{\bf{ = 0}}{\bf{.003}}\)

Mole Fraction of \({{\bf{H}}_{\bf{2}}}{\bf{O = }}\frac{{{\bf{5}}{\bf{.66}}}}{{{\bf{5}}{\bf{.66 + 0}}{\bf{.0147}}}}{\bf{ = 0}}{\bf{.997}}\)

Mole Fraction of Codeine and Ethanol

Mass of \({{\bf{C}}_{{\bf{18}}}}{{\bf{H}}_{{\bf{21}}}}{\bf{N}}{{\bf{O}}_{\bf{3}}}\)= 46.85g

Molar Mass of \({{\bf{C}}_{{\bf{18}}}}{{\bf{H}}_{{\bf{21}}}}{\bf{N}}{{\bf{O}}_{\bf{3}}}\) = 299.4 g/mole

Number of Moles of \({{\bf{C}}_{{\bf{18}}}}{{\bf{H}}_{{\bf{21}}}}{\bf{N}}{{\bf{O}}_{\bf{3}}}{\bf{ = }}\frac{{{\bf{46}}{\bf{.85}}}}{{{\bf{299}}{\bf{.4}}}}{\bf{ = 0}}{\bf{.156}}\)

Mass of Ethanol = 125.5g

Molar Mass of Ethanol = 46 g/mole

Number of Moles of \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH = }}\frac{{{\bf{125}}{\bf{.5}}}}{{{\bf{46}}}}{\bf{ = 2}}{\bf{.73}}\)

Mole Fraction of \({{\bf{C}}_{{\bf{18}}}}{{\bf{H}}_{{\bf{21}}}}{\bf{N}}{{\bf{O}}_{\bf{3}}}{\bf{ = }}\frac{{{\bf{0}}{\bf{.156}}}}{{{\bf{0}}{\bf{.156 + 2}}{\bf{.73}}}}{\bf{ = 0}}{\bf{.054}}\)

Mole Fraction of \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH = }}\frac{{{\bf{2}}{\bf{.73}}}}{{{\bf{2}}{\bf{.73 + 0}}{\bf{.156}}}}{\bf{ = 0}}{\bf{.945}}\)

Mole Fraction of I2 and Ethanol

Mass of I₂= 25g

Molar Mass of \({{\bf{I}}_{\bf{2}}}{\bf{ = 253}}{\bf{.81 g/mole}}\)

Number of Moles of \({{\bf{I}}_{\bf{2}}} = \frac{{25}}{{253.81}} = 0.098\)

Mass of Ethanol = 125g

Molar Mass of Ethanol = 46 g/mole

Number of Moles = \(\frac{{125}}{{46}} = 2.72\)

Mole Fraction of \({{\bf{I}}_{\bf{2}}}{\bf{ = }}\frac{{{\bf{0}}{\bf{.098}}}}{{{\bf{0}}{\bf{.098 + 2}}{\bf{.72}}}}{\bf{ = 0}}{\bf{.035}}\)

Mole Fraction of \({\bf{Ethanol = }}\frac{{{\bf{2}}{\bf{.72}}}}{{{\bf{2}}{\bf{.72 + 0}}{\bf{.098}}}}{\bf{ = 0}}{\bf{.96}}\)