2. A solution of calcium chloride was made by dissolving 2.77 grams in 100.0 mL
ID: 575303 • Letter: 2
Question
2. A solution of calcium chloride was made by dissolving 2.77 grams in 100.0 mL of deionized water. This solution was labeled "Solution # 1 then made The following dilutions were 10.0 mL Solution #1 was diluted to 500.0 mL and labeled "Solution # 2". 5.0 mL of Solution # 2 was diluted to 125.00 mL and labeled "Solution # 3" 1.25 mL of Solution # 3 was diluted to 20.0 mL and labeled "Solution #4" What were the concentrations, in molarity, of solutions #1-4? How many the end of the serial dilution? a. b. micrograms of calcium chloride were present in 5.0 mL of solution # 4 atExplanation / Answer
Given:
2.77 grams of CaCl2 is dissolved in 100.0ml of deionized water. This is labeled as "Solution #1".
We know that:
1. Number of moles of solute = Weight of Solute / Molecular Weight of Solute
2. Molarity = Number of moles of solute / Volume of solution( in Litres)
3. Molecular Weight of CaCl2 is 110 gram/mole
Solution:
Number of moles of Solute present in Solution #1 = Weight of CaCl2 / Molecular Weight of CaCl2
= 2.77/110
= 0.025 mole
Concentration of Solution #1 = No. of moles of CaCl2 in Solution #1 / Volume of Solution(in Litres)
= 0.025/0.1
= 0.25M
Now,
no. of moles present in 10 ml of Solution #1 = Concentration of Solution#1 * Given Volume
= 0.25*0.01
= 0.0025 moles of CaCl2
As 10.0 ml of Solution #1 is diluted to 500 ml to make Solution #2, the Number of moles in Solution #2 is same as that of the 10 ml Solution #1.
So, Number of moles of CaCl2 present in the Solution #2 = 0.0025 moles
Concentration of Solution #2= No. of moles of CaCl2in Solution #2 / Volume of Solution(in Litres)
= 0.0025/0.5
= 0.005M
no. of moles of CaCl2 present in 5 ml of Solution #2 = Concentration of Solution#2 * Given Volume
= 0.005*5*10-3
= 25*10-6 moles
As 5 ml of Solution #2 is diluted to 125 ml to make Solution #3, the Number of moles in Solution #3 is same as that of the 5 ml Solution #2.
So, Number of moles of CaCl2 present in the Solution #3 = 25*10-6 moles
Concentration of Solution #3 = No. of moles of CaCl2 in Solution #3 / Volume of Solution(in Litres)
= (25*10-6)/0.125
= 2*10-4M
no. of moles of CaCl2 present in 1.25 ml of Solution #3 = Concentration of Solution#3 * Given Volume
= 2*10-4 * 1.25*10-3
= 2.5*10-7moles
As 1.25 ml of Solution #3 is diluted to 20 ml to make Solution #4, the Number of moles in Solution #4 is same as that of the 1.25 ml Solution #3.
So, Number of moles of CaCl2 present in the Solution #4 = 2.5*10-7moles
Concentration of Solution #4 = No. of moles of CaCl2 in Solution #4 / Volume of Solution(in Litres)
= 2.5*10-7/0.02
=1.25*10-5M
The Concentrations are 0.25M,0.005M,2*10-4M,1.25*10-5M for Solution#1, Solution#2, Solution#3, Solution#4 respectively. This answers for the Part A.
When 5ml is taken out from Solution #4, no .of moles of CaCl2 present in it is
no. of moles of CaCl2 present in 5ml of Solution #4= Concentration of Solution#4 * Given Volume
= 5*10-3*1.25*10-5
= 6.25*10-8moles
Number of moles of solute = Weight of Solute / Molecular Weight of Solute
On tranposing the above equation, we get
Weight of Solute= Number of moles of solute*Molecular Weight of Solute
Therefore Weight of CaCl2= 6.25*10-8*110
= 6.874 Micrograms
This answers the part B
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