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Glitzey
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0.502 g of an alkali metal sulfate is dissolved in water and excess barium chloride solution,
BaCl aq) 2(aq), is added to precipitate all the sulfate ions as barium sulfate, BaSO4 (s). The
precipitate is filtered and dried and weighs 0.672 g.

Determine the molar mass of the alkali metal sulfate and state its units. [2 marks]
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charco
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(Original post by Glitzey)
0.502 g of an alkali metal sulfate is dissolved in water and excess barium chloride solution,
BaCl aq) 2(aq), is added to precipitate all the sulfate ions as barium sulfate, BaSO4 (s). The
precipitate is filtered and dried and weighs 0.672 g.

Determine the molar mass of the alkali metal sulfate and state its units. [2 marks]
Use the info in the question to calculate the number of moles of BaSO4 formed = 0.672/Mr(BaSO4) = 0.672/233

The equation for the reaction is:

M2SO4 + BaCl2 --> BaSO4 + 2MCl

in other words 1 mole of M2SO4 makes 1 mole of BaSO4

Thus moles of M2SO4 = 0.672/233 = 0.00288

You know the mass of the M2SO4 = 0.502 g

Therefore the Mr = 0.502/0.00288 = 174
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ragnar_jonsson
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The full question:
(a) Calculate the amount (in mol) of barium sulfate formed. [2]
(b) Determine the amount (in mol) of the alkali metal sulfate present. [1]
(c) Determine the molar mass of the alkali metal sulfate and state its units. [2]
(d) Deduce the identity of the alkali metal, showing your working. [2]
(e) Write an equation for the precipitation reaction, including state symbols. [2]

X is an alkali metal (Period 1)
(SO4)2- is an anion with charge -2
thus the balanced compound is X2SO4
2X + (SO4)2- --> X2SO4

mass of X2SO4 = 0.502 g

a) moles of BaSO4?
Precipitate Mass = 0.672 g
M = 233.4 g mol-1
n = 0.00288 mol

b) X2SO4 + BaCL2 --> BaSO4 + 2 XCL
n(X2SO4)=n(BaSO4)
n=0.00288 mol

c) Molar mass of X2SO4?
m/n = M
0.502 g/0.00288 mol = 174.3 g/mol

d) 2 M(X) + 1 M(S) + 4 M(O) = 174.3 g/mol
2 M(X) = 78.3 g/mol
M(X) = 39.15 g/mol
the alkali metal is Potassium/ K (M=39.10 g/mol)

e) K2SO4 (aq) + BaCl2 (aq) --> BaSO4 (s) + 2 KCl (aq)

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Sharod100
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(Original post by ragnar_jonsson)
The full question:
(a) Calculate the amount (in mol) of barium sulfate formed. [2]
(b) Determine the amount (in mol) of the alkali metal sulfate present. [1]
(c) Determine the molar mass of the alkali metal sulfate and state its units. [2]
(d) Deduce the identity of the alkali metal, showing your working. [2]
(e) Write an equation for the precipitation reaction, including state symbols. [2]

X is an alkali metal (Period 1)
(SO4)2- is an anion with charge -2
thus the balanced compound is X2SO4
2X + (SO4)2- --> X2SO4

mass of X2SO4 = 0.502 g

a) moles of BaSO4?
Precipitate Mass = 0.672 g
M = 233.4 g mol-1
n = 0.00288 mol

b) X2SO4 + BaCL2 --> BaSO4 + 2 XCL
n(X2SO4)=n(BaSO4)
n=0.00288 mol

c) Molar mass of X2SO4?
m/n = M
0.502 g/0.00288 mol = 174.3 g/mol

d) 2 M(X) + 1 M(S) + 4 M(O) = 174.3 g/mol
2 M(X) = 78.3 g/mol
M(X) = 39.15 g/mol
the alkali metal is Potassium/ K (M=39.10 g/mol)

e) K2SO4 (aq) + BaCl2 (aq) --> BaSO4 (s) + 2 KCl (aq)

What Exam is this question from?
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charco
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(Original post by Sharod100)
What Exam is this question from?
IB HL November 2007
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Mandela123
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How is the number of moles of the alkali metal sulfate equals to that of BaSO4 dont get it
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charco
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(Original post by Mandela123)
How is the number of moles of the alkali metal sulfate equals to that of BaSO4 dont get it
look at the equation:

1 sulfate ion on the LHS and 1 on the RHS
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