Chapter 19, Problem 17STP

Interpretation Introduction

Interpretation:

The flask containing 9.32 g chemical out of the given five chemicals is to determined.

Concept introduction:

The question is based on the concept of molarity.

Molarity is ratio of number of moles of solute to the volume of solution in liters.

$Molarity=\frac{numberofmolesofsolute}{volumeofsolutioninliters}$

Molarity is expressed in $mol/L$

Further, $\begin{array}{l}Numberofmoles=\raisebox{1ex}{$givenmass$}\!\left/ \!\raisebox{-1ex}{$molarmass$}\right.\\ \therefore Molarity=\frac{givenmassofsolute}{molarmassofsolute\times volumeofsolutioninliters}\\ Molarity=\frac{m}{M\times V}\\ where\\ misgivenmassofsolute\\ Mismolarmassofsolute\\ Visvolumeofsolutioninliters\\ Therefore,\\ m=molarity\times M\times V\end{array}$

Answer to Problem 17STP

A

Explanation of Solution

Molar mass of each substance is calculated as follows:

$\begin{array}{l}{M}_{KCl}=39+35.5\\ =74.5g/mol\\ M_{C{H}_{3}OH}=12+3\times 1+16+1\\ =32g/mol\\ M_{Ba{\left(OH\right)}_2}=137+2\times \left(16+1\right)\\ =171g/mol\\ M_{C{H}_{3}COOH}=12\times 2+16\times 2+1\times 4\\ =60g/mol\\ M_{NaOH}=23+16+1\\ =40g/mol\end{array}$

Mass of each sunstance can be calculated as follows:

$\begin{array}{l}{m}_{KCl}=74.5\times 0.25\times 0.5\\ =9.31g\\ m_{C{H}_{3}OH}=32\times 0.25\times 0.5\\ =4g\\ m_{Ba{\left(OH\right)}_2}=171\times 0.25\times 0.5\\ =21.4g\\ m_{C{H}_{3}COOH}=60\times 0.25\times 0.5\\ =7.5g\\ m_{NaOH}=40\times 0.25\times 0.5\\ =5g\end{array}$

Conclusion

Mathematical calculations show that option A is correct.