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Chapter 3: Q62 E (page 171)

What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-M solution is allowed to evaporate until the volume is reduced to 0.105 L?

Short Answer

Expert verified

The resulting concentration after evaporation is 0.794M.

Step by step solution

01

Given data

The number of moles of solute before dilution is equal to the number of moles after dilution.

\({M_1}{V_1} = {M_2}{V_2}\).

where

\(\begin{align}{M_1} &= \,Concentration\,of\,stock\,solution\\{V_1} &= \,Volume\,of\,stock\,solution\,\\{M_2} &= \,Concentration\,of\,the\,new\,solution\\{V_2} &= \,Final\,volume\,of\,the\,solution\end{align}\)

Given

\(\begin{align}{M_1} = 0.556M\\{V_2} = 0.105L\\{V_1} = 0.150L\end{align}\)

02

Determine the final concentration

Find \({M_2}\), the final concentration.

\({M_2} = \frac{{{M_1}{V_1}}}{{{V_2}}}\)

Substitute the values in the formula above.

\({M_2} = \frac{{\left( {0.556M} \right)\left( {0.150L} \right)}}{{0.105L}} = 0.794M\)

Therefore, the resulting concentration of NaCl after evaporation is 0.794M.

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