Question

The density of a $10\%$by mass of $KCl$solution in water is $1.06$ $g$ $m{L^{ - 1}}$. Find out the mole fraction, molarity and molality of $KCl$ in this solution respectively.
(i) $0.026;$ $1.42$ $M;$ $1.489$ $m$
(ii) $0.026;$ $2.42$ $M;$ $1.489$ $m$
(iii) $0.036;$ $2.42$ $M;$ $1.589$ $m$
(iv) $0.036;$ $1.42$ $M;$ $1.589$ $m$

Answer 1

$x\%$ by mass of any solute means in $100$ $g$ of the solution $x$ $g$ of solute is present. So by applying the formula of density we can calculate the volume of the given solution. Also we can calculate the number of moles of solvent and solute present in the solution as the molecular weights are known hence the mole fraction can be calculated. Then by using the formula of molarity and molality we can calculate them accordingly.

Complete step-by-step solution: $10\%$ by mass of $KCl$ solution means in $100$ $g$ of the solution $10$ $g$ of $KCl$is present.
Also it is given that the density of the $KCl$solution in water is $1.06$ $g$ $m{L^{ - 1}}$.
We know, $Density\, = \,\dfrac{{Mass}}{{Volume}}$
$\Rightarrow$ $Volume\, = \,\dfrac{{Mass}}{{Density}}$
For the given $KCl$solution density is $1.06$ $g$ $m{L^{ - 1}}$ and the mass is $100$ $g$.
$\therefore$ $Volume\, = \,\dfrac{{100\,g}}{{1.06\,g\,m{L^{ - 1}}}}$
$\Rightarrow$ $Volume$ $=$ $94.34$ $mL$
Therefore the volume of the solution is $94.34$ $mL$.
Now, number moles of a compound $= \,\dfrac{{Given\,Mass}}{{Molar\,Mass}}$
Molar Mass of KCl$$$ = $$ 74.5$$g$$mo{l^{ - 1}}$Given Mass of$KClin the solution$$ = $$ 10,g$$\therefore ,$number of moles of$KCl$in the solution$ = ,\dfrac{{10,g}}{{74.5,g,mo{l^{ - 1}}}}, = ,0.134,mol$Molar Mass of${H_2}O$$$ = $18$ $g$ $mo{l^{ - 1}}$ Given mass of ${H_2}O$in the solution = $$ $($Mass of solution $-$ Mass of $KCl$in solution $)$ $g$
$=$ $\left( {100 - 10} \right)\,g\, = \,90\,g$
$\therefore \,$number of moles of ${H_2}O$in the solution $= \,\dfrac{{90\,g}}{{18\,g\,mo{l^{ - 1}}}}\, = \,5\,mol$
Now, mole fraction of a compound in a solution$=\,\dfrac{{Number\,of\,moles\,of\,the\,compound\,present\,in\,solution}}{{Total\,number\,of\,moles\,present\,in\,the\,solution}}$
Total number of moles present in solution $= \,($Number of moles of $KCl$ present in solution$+$ Number of moles of ${H_2}O$ present in solution$)$ $mol$ $=$ $\left( {0.134 + 5} \right)\,mol\, = \,5.134\,mol$
Therefore, mole fraction of $KCl$ in solution $= \,\dfrac{{0.134\,mol}}{{5.134\,mol}}\, = \,0.026$
Now,Molarity=$\,\dfrac{{Number\,of\,moles\,of\,solute\,present\,in\,solution}}{{Volume\,of\,solution\,\left( {in\,L} \right)}}$
In the given solution the solute is $KCl$.
Therefore, Molarity $= \,\dfrac{{0.134\,mol}}{{93.34\,mL}}\, = \,\dfrac{{0.134\,mol}}{{94.34 \times {{10}^{ - 3}}L}}\, = \,\,1.42\,M$ $\left( {\because \,1\,L\, = \,1000\,mL} \right)$
Molality $= \,\dfrac{{Number\,of\,moles\,of\,solute\,present\,in\,solution}}{{Mass\,of\,solvent\,\left( {in\,kg} \right)}}$
In the given solution the solute is $KCl$ and the solvent is ${H_2}O$ .
Therefore, Molality $= \,\dfrac{{0.134\,mol}}{{90\,g}}\, = \,\dfrac{{0.134\,mol}}{{90 \times {{10}^{ - 3}}\,kg}}\, = \,\,1.489\,m$ $\left( {\because \,1\,kg\, = \,1000\,g} \right)$

Hence the correct answer is (i) $0.026;$ $1.42$ $M;$ $1.489$ $m$.

Note: Identify the solvent and solute properly otherwise it will lead wrong calculations. Take proper care of the units. Try to write the units in each step so that mistakes can be avoided. Also check which unit is being used in a particular formula and hence change the units wherever necessary.