Question

A solution containing \[10.0\] millimoles of \[{\text{CaC}}{{\text{l}}_2}\] is diluted to 1 L. The number of grams of \[{\text{CaC}}{{\text{l}}_2}.2{{\text{H}}_2}{\text{O}}\] per mL of the final solution is \[x \times {10^{ - 3}}{\text{g m}}{{\text{L}}^{ - 1}}.\] Then, the nearest integer of \[2x\] is:

Answer 1

Hint: In this question we have to calculate the amount of \[{\text{CaC}}{{\text{l}}_2}\] per milliliters. Then we can compare with the given equation to final the value of x and 2x further.

Formula used:
\[{\text{concentration}} = \dfrac{{{\text{number of moles }}}}{{{\text{volume in liter}}}}\]

Complete step by step solution:
We have been given the millimoles of \[{\text{CaC}}{{\text{l}}_2}\]and volume of \[{\text{CaC}}{{\text{l}}_2}\]. Let us calculate the concentration of \[{\text{CaC}}{{\text{l}}_2}\]:
The suffix mili used before moles is for \[{10^{ - 3}}\]. So 10 millimoles of \[{\text{CaC}}{{\text{l}}_2}\] is equal to \[10 \times {10^{ - 3}} = {10^{ - 2}}\] moles.
And the total volume is given 1L. So:
\[{\text{concentration of CaC}}{{\text{l}}_2} = \dfrac{{{\text{number of moles of CaC}}{{\text{l}}_2}{\text{ }}}}{{{\text{volume of CaC}}{{\text{l}}_2}}}\]
\[ \Rightarrow {\text{concentration}} = \dfrac{{{{10}^{ - 2}}{\text{ moles}}}}{{1{\text{ L}}}}\]
\[ \Rightarrow {10^{ - 2}}{\text{ mole }}{{\text{L}}^{ - 1}}\]
But we need the concentration in \[{\text{g m}}{{\text{L}}^{ - 1}}\] . So, we have to convert moles in mass by multiplying with molar mass:
Molar mass of Mass \[{\text{CaC}}{{\text{l}}_2}.2{{\text{H}}_2}{\text{O}}\] is \[40 + 35.5 \times 2 + 2(2 \times 2 + 16) = 147{\text{ g mo}}{{\text{l}}^{ - 1}}\]
Replacing mole by \[147{\text{ g mo}}{{\text{l}}^{ - 1}}\] , we will get:
\[{\text{concentration of CaC}}{{\text{l}}_2}.2{{\text{H}}_2}{\text{O = 147 }} \times {\text{1}}{{\text{0}}^{ - 2}}{\text{ g}}{{\text{L}}^{{\text{ - 1}}}}\]
\[ \Rightarrow {\text{concentration}} = 1.47{\text{ g }}{{\text{L}}^{ - 1}}\]
Since 1 liter is equal to 1000 milliliter, we will divide the above value with 1000, which gives:
\[{\text{concentration of CaC}}{{\text{l}}_2}.2{{\text{H}}_2}{\text{O = 1}}{\text{.47}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ g m}}{{\text{L}}^{ - 1}}\]
We have been given amount of \[{\text{CaC}}{{\text{l}}_2}.2{{\text{H}}_2}{\text{O}}\] in question as \[x \times {10^{ - 3}}{\text{g m}}{{\text{L}}^{ - 1}}.\]
Comparing both of the above equation, we will get:
\[x \times {10^{ - 3}}{\text{g m}}{{\text{L}}^{ - 1}} = 1.47 \times {10^{ - 3}}{\text{g m}}{{\text{L}}^{ - 1}}\]
$ \Rightarrow x = 1.47$
But we have to calculate the nearest integer to 2x which will be:
\[2x = 2 \times 1.47\]
Solving this for x, we get:
\[x = 2.94 \simeq 3\]

Hence the answer is 3.

Note: Calcium chloride is a white, crystalline salt and very highly soluble in water. It is formed by the neutralization reaction of hydrochloric acid and calcium hydroxide, that is it is a salt formed by strong acid that is hydrochloric acid and a string base that is calcium hydroxide. It is a non volatile solute and when added to water, it relatively lowers the vapor pressure of the water. It is used in deicing or depression in freezing point, in road surfacing and as an additive in food.