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Question: A solution containing \[10.0\] millimoles of \[{\text{CaC}}{{\text{l}}_2}\] is diluted to 1 L. The n...

A solution containing 10.010.0 millimoles of CaCl2{\text{CaC}}{{\text{l}}_2} is diluted to 1 L. The number of grams of CaCl2.2H2O{\text{CaC}}{{\text{l}}_2}.2{{\text{H}}_2}{\text{O}} per mL of the final solution is x×10−3g mL−1.x \times {10^{ - 3}}{\text{g m}}{{\text{L}}^{ - 1}}. Then, the nearest integer of 2x2x is:

Explanation

Solution

In this question we have to calculate the amount of CaCl2{\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:
concentration=number of moles volume in liter{\text{concentration}} = \dfrac{{{\text{number of moles }}}}{{{\text{volume in liter}}}}

Complete step by step solution:
We have been given the millimoles of CaCl2{\text{CaC}}{{\text{l}}_2}and volume of CaCl2{\text{CaC}}{{\text{l}}_2}. Let us calculate the concentration of CaCl2{\text{CaC}}{{\text{l}}_2}:
The suffix mili used before moles is for 10−3{10^{ - 3}}. So 10 millimoles of CaCl2{\text{CaC}}{{\text{l}}_2} is equal to 10×10−3=10−210 \times {10^{ - 3}} = {10^{ - 2}} moles.
And the total volume is given 1L. So:
concentration of CaCl2=number of moles of CaCl2 volume of CaCl2{\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}}}
⇒concentration=10−2 moles1 L\Rightarrow {\text{concentration}} = \dfrac{{{{10}^{ - 2}}{\text{ moles}}}}{{1{\text{ L}}}}
⇒10−2 mole L−1\Rightarrow {10^{ - 2}}{\text{ mole }}{{\text{L}}^{ - 1}}
But we need the concentration in g mL−1{\text{g m}}{{\text{L}}^{ - 1}} . So, we have to convert moles in mass by multiplying with molar mass:
Molar mass of Mass CaCl2.2H2O{\text{CaC}}{{\text{l}}_2}.2{{\text{H}}_2}{\text{O}} is 40+35.5×2+2(2×2+16)=147 g mol−140 + 35.5 \times 2 + 2(2 \times 2 + 16) = 147{\text{ g mo}}{{\text{l}}^{ - 1}}
Replacing mole by 147 g mol−1147{\text{ g mo}}{{\text{l}}^{ - 1}} , we will get:
concentration of CaCl2.2H2O = 147 ×10−2 gL - 1{\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}}}}
⇒concentration=1.47 g L−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:
concentration of CaCl2.2H2O = 1.47×10−3 g mL−1{\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 CaCl2.2H2O{\text{CaC}}{{\text{l}}_2}.2{{\text{H}}_2}{\text{O}} in question as x×10−3g mL−1.x \times {10^{ - 3}}{\text{g m}}{{\text{L}}^{ - 1}}.
Comparing both of the above equation, we will get:
x×10−3g mL−1=1.47×10−3g mL−1x \times {10^{ - 3}}{\text{g m}}{{\text{L}}^{ - 1}} = 1.47 \times {10^{ - 3}}{\text{g m}}{{\text{L}}^{ - 1}}
⇒x=1.47\Rightarrow x = 1.47
But we have to calculate the nearest integer to 2x which will be:
2x=2×1.472x = 2 \times 1.47
Solving this for x, we get:
x=2.94≃3x = 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.