Question

A solution contains 0.115 mol

H2OH_2O

and an unknown number of moles of sodium ch lo ride. The vapor pressure of the solution at

30C30 ^ { \circ } \mathrm { C }

is 25.7 torr. The vapor pressure of pure water at this temperature is 31.8 torr. Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)

Solutions

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The number of moles of water is nH2O=0.115moln_{H_2O} = 0.115 \mathrm{mol}

The vapor pressure of the sodium chloride solution at 30C30^{\circ} \mathrm{C} is Psolution=25.7torrP_{solution} = 25.7 \mathrm{torr}

The vapor pressure of pure water is PH2Oo=31.8torrP^{o}_{H_2O} = 31.8 \mathrm{torr}

Let us calculate the mass of sodium chloride in the solution.

First, we have to find the mole fraction of water.

Psolution=XH2OPH2OoXH2O=PsolutionPH2Oo=25.7torr31.8torr=0.808\begin{align*} P_{solution} &= X_{H_2O} \cdot P^{o}_{H_2O} \\ X_{H_2O} &= \frac { P_{solution} } { P^{o}_{H_2O} }\\ &= \frac { 25.7 \mathrm{torr} } { 31.8 \mathrm{torr} }\\ &= 0.808 \end{align*}

Now, we can calculate the number of moles of NaCl\mathrm{NaCl}.

  • Since NaCl\mathrm{NaCl} is strong electrolyte, it is completely ionized to form 2 mol\mathrm{mol} of dissolved particles. Therefore

XH2O=nH2OnH2O+2nNaCl0.808=0.115mol0.115mol+2nNaCl0.808(0.115mol+2nNaCl)=0.115mol0.093mol+1.616nNaCl=0.115mol1.616nNaCl=0.022molnNaCl=0.0136mol\begin{align*} X_{H_2O} &= \frac { n_{H_2O} } { n_{H_2O} + 2 \cdot n_{NaCl} }\\ 0.808 &= \frac {0.115 \mathrm{mol} } { 0.115 \mathrm{mol} + 2 \cdot n_{NaCl} }\\ 0.808 \cdot (0.115 \mathrm{mol} + 2 \cdot n_{NaCl}) &= 0.115 \mathrm{mol} \\ 0.093 \mathrm{mol} + 1.616 \cdot n_{NaCl} &= 0.115 \mathrm{mol} \\ 1.616 \cdot n_{NaCl} &= 0.022 \mathrm{mol}\\ n_{NaCl} &= 0.0136 \mathrm{mol} \end{align*}

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