Question

Which of the following have equal mass of $C{l^ - }$ions in $1.0L$of each of the following solutions?
A.$5\% NaCl$(density=$1.07g/ml$)
B.$5\% KCl$(density=$1.06g/ml$)
C.58.5 g $NaCl$
D.55.5g $BaC{l_2}$

Answer 1

Hint:In solution, mass concentration is commonly encountered as the ratio of mass/volume. Molarity, Molality, normality are various methods by which we can find out various quantities required. Moreover, for a pure chemical, the mass concentration is equal to its density, thus the mass concentration of a component in a mixture can be called the density of that component in the mixture.
Formula used:
$M = \dfrac{{\% byweight \times 10 \times d}}{{M{w_2}}}$
Where d =density
$M{w_2}$=molecular weight of compound

Complete step by step answer:
Now, we will solve each and every option and then determine the answer.
Firstly,
Mass of $C{l^ - }$ is $35.5g$
Now, using the above formula,
Density $ = 1.07g/ml$
$M{w_2} = 58.5$
Percentage$ = 5\% $
$M = \dfrac{{5 \times 10 \times 1.07}}{{58.5}} = 0.91M$( $C{l^ - } = 0.91$,$M = 0.91 \times 35.5 = 32.4g$)
Secondly,
Density $ = 1.06g/ml$
Percentage$ = 5\% $
$M = \dfrac{{5 \times 10 \times 1.06}}{{74.5}} = 0.77M$($C{l^ - } = 0.77,M = 0.77 \times 35.5 = 25.25g$)
Now, in the third case,
Weight of solid$ = 58.5g$
$M = \dfrac{{{W_2} \times 100}}{{M{w_2} \times V}}$
$ = \dfrac{{58.5 \times 1000}}{{58.5 \times 1000}} = 1M$ ($C{l^ - } = 1M = 1 \times 35.5 = 35.5g$
Now, in the last case, the weight of the compound is $55.5g$
Weight of solid = 55.5g
$M = \dfrac{{55.5 \times 1000}}{{111 \times 1000}} = 0.5M$
$BaC{l_2} \to B{a^{2 + }} + 2C{l^ - }$
Therefore, $C{l^ - } = 2 \times M = 2 \times 0.5 = 1M$
$ = 1 \times 35.5 = 35.5g$

Hence, option C and D are correct.

Note:
There are two types of ions- cations and anions. A cation is a positively charged ion and is formed by the loss of one or more electrons by an atom whereas an anion is a negatively charged ion and is formed by gaining one or two electrons from the atom.