Multiconcept Problem 12.163 Part A When 22.79 g of a mixture of an alkali metal
ID: 577975 • Letter: M
Question
Multiconcept Problem 12.163 Part A When 22.79 g of a mixture of an alkali metal chloride (XCI) and an akaline earth chlcride (YCl,) was dissolved in 350.0 g of water, te freang point of the resultant solution was 432-C. Addition of an excess of aqueous AgNO, to the solution yielded How much of each metal chloride was present in the original miue? Express your answers using three significant figures separated by a comma to m(xCi), m(YCh)- 4.71,18.08 Submit Previous Amwers Reqest Anewer X Incorrect; Try Again; 5 attempts remaining -Part B What are the identities of the two metals X and Y Express your answer as chemical symbols separated by a comma Submit Previous Answers Reavest Answr Incorrect; Try Again; 5 attempts remaining Provide FeedbackExplanation / Answer
Ans. Step 1: Let the moles of XCl and YCl2 taken be A and B moles, respectively.
# Given, mass of white precipitate (AgCl) = 66.664 g
Moles of AgCl = Mass/ Molar mass = 66.664 g / (143.3209 g/ mol) = 0.46514 mol
# Since 1 mol AgCl consists of 1 mol Cl, the total number of moles of Cl in AgCl must be equal to the sum of moles of Cl in XCl and YCl2.
# 1 mol XCl yields 1 mol Cl. 1 mol YCl2 yields 2 mol Cl.
So, total number of moles of Cl in sample = 1 x moles of XCl + 2 x moles of YCl2
= 1 x A mol + 2 x B mol
= (A + 2B) mol
Now,
(A + 2B) mol = 0.46514 mol
Hence, A + 2B = 0.46514 - equation 1
# Step 2: Molality of XCl in solution = Moles of XCl / Mass of solvent in kg
= A mol / 0.350 kg
= 2.85714A m
Molality of YCl2 in solution = B mol / 0.350 kg = 2.85714B m
# Depression in freezing point of the solution is given by-
dTf = i Kf m
where, i = Van’t Hoff factor. [i = 2 for XCl ; i = 3 for YCl2
Kf = molal freezing point depression constant of solvent = 1.860C / m
m = molality of the solution
dTf = Freezing point of pure solvent – Freezing point of solution
Total depression in freezing point is equal to the depression in freezing point due to XCl and YCl2.
Putting the values in above expression-
0.00C – (-4.320C) = (2 x 1.860C m-1 x 2.85714A m) + (3 x 1.860C m-1 x 2.85714B m)
Or, 4.320C = (1.860C m-1 x 2.85714 m) (2A + 3B)
Or, 4.320C / 5.31428040C = 2A + 3B
Hence, 2A + 3B = 0.81290 - equation 2
# Step 3: Comparing (equation 1 x 2) – equation 2-
2A + 4B = 0.93028
(-) 2A + 3B = 0.81290
-------------------------------
B = 0.11738
Therefore, moles of YCl2 taken = B mol= 0.11738 mol
# Putting the value of B in equation 1-
A + 2 (0.11738) = 0.46514
Or, A = 0.46514 - 0.23476
Hence, A = 0.23038
Therefore, moles of XCl taken = A mol = 0.23038 mol
# Step 4: Let the molar mass of X and Y be M g/mol and N g/mol, respectively.
Total mass of sample = (Moles of XCl x MW) + (Moles of YCl2 x MW)
Or, 22.79 g = [ 0.23038 mol x {M g mol-1 + (1 x 35.4527 g mol-1)}] +
[ 0.11738 mol x M g mol-1 x (2 x 35.4527 g mol-1)]
Or, 22.79 = (0.23038M + 8.167593026) + (0.11738N + 8.322875852)
Or, 22.79 = 0.23038M + 0.11738N + 16.490468878
Or, 22.79 - 16.490468878 = 0.23038M + 0.11738N
Hence, 0.23038M + 0.11738N = 6.2995 - equation 4
# We don’t enough data to correlate further. So, we need to use “hit & trial” method to fit the elements from group I and II to see if the sum of masses 0.23038M + 0.11738N = 6.2995) becomes equal to 22.79 gram.
It is- X = N ; Y = Be
Therefore, XCl = NaCl ; YCl2 = BeCl2
# Now, mass of XCl = 0.23038 mol x (58.442468 g/mol) = 13.464 g
Mass of YCl2 = 22.79 – 13.464 g = 9.326 g
Note: The mass of XCl and YCl2 may exhibit a bit deviation from actual value of summed mass 22.79 g due to accounting significant figures.
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