A 0.40 M solution of calcium chloride is diluted by a factor of 4. What is the c
ID: 514304 • Letter: A
Question
A 0.40 M solution of calcium chloride is diluted by a factor of 4. What is the concentration of the chloride ion in the resulting solution? a. 0.10 M Cl^- b. 0.20 M Cl^- c. 0.40 M Cl^- d. 0.60 M Cl^- e. 0.80 M Cl^- If 20.0 mL of a solution that is 12%w/v is diluted with water until the volume is 100.0 mL, what is the concentration of the new solution? a. 0.024%w/v b. 0.12%w/v c. 2.0%w/v d. 2.4%w/v e. 60 %w/v All compounds shown are soluble water. If 1.5 M aqueous solutions were prepared pure using each of these compounds, which solution's freezing point of pure water? a. NaCl b. CaCl_2 c. lithium nitrate d. potassium sulfate e. sucrose, C_12H_22O_11Explanation / Answer
initial molarity of CaCl2 = 0.40M that is 0.40moles of CaCl2 in 1 L of solution
It is diluted by 4 times.
Now the new solution has
moles of solute = 0.4 moles
volume of solution = 4 L
Thus the molarity of diluted solution = 0.4/4 = 0.1 M
we need [Cl-] in this solution.
Since once Ca Cl2 has 2 Cl- ions
the molarity of [Cl-] in diluted solution = 2 x0.1 = 0.20 M
20) initial volume = 20 mL and strength = 12% w/v
now dilluted to 100mL and strength = 12 x20/100 = 2.4%
21) Each of the solution has same molarity = 1.5M
We know depression in freezing point = i x m x Kf
where i = van't Hoff factor , m = molaloty and Kf is the molar depression constant of the solvent
considering M = m,
all have m same , Kf same, then delta Tf is proprtional to i .
The salt with lowest i value has lowest delta Tf and has freezing point closer to that of pure water.
NaCl , i=2 (NaCl -----> Na+ +Cl-)
CaCl2 ,i= 3 (CaCl2 ------> Ca+2 + 2Cl-)
LiNO3 , i=2 (LiNO3 ---------->Li+ + NO3-)
K2SO4 , i= 3 (K2SO4 ------> 2K+ + SO4-2)
sucrose i=1 (no ions)
Thus the freezing point of sucrose solution will be closer to that of water
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