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4 g equimolar mixture of NaOH and Na2CO3 contains x g of NaOH and y g of Na2CO3. The value of x is _________ g. (Nearest integer)
Correct answer is '1'. Can you explain this answer?
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4 g equimolar mixture of NaOH and Na2CO3 contains x g of NaOH and y g ...
Solution:
Given, 4 g mixture of NaOH and Na2CO3.
Let the molar mass of NaOH and Na2CO3 be M1 and M2 respectively.
Let the number of moles of NaOH and Na2CO3 be n1 and n2 respectively.
Then, n1 = x/M1 and n2 = y/M2
As the mixture is equimolar, n1 = n2
Therefore, x/M1 = y/M2
Also, x + y = 4
On solving these two equations, we get
x = 2M1/(M1 + M2)
As NaOH and Na2CO3 have molar masses of 40 g/mol and 106 g/mol respectively,
x = 2*40/(40 + 106) = 1.01 ≈ 1 g (nearest integer)
Therefore, the value of x is 1 g.
Given, 4 g mixture of NaOH and Na2CO3.
Let the molar mass of NaOH and Na2CO3 be M1 and M2 respectively.
Let the number of moles of NaOH and Na2CO3 be n1 and n2 respectively.
Then, n1 = x/M1 and n2 = y/M2
As the mixture is equimolar, n1 = n2
Therefore, x/M1 = y/M2
Also, x + y = 4
On solving these two equations, we get
x = 2M1/(M1 + M2)
As NaOH and Na2CO3 have molar masses of 40 g/mol and 106 g/mol respectively,
x = 2*40/(40 + 106) = 1.01 ≈ 1 g (nearest integer)
Therefore, the value of x is 1 g.
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Community Answer
4 g equimolar mixture of NaOH and Na2CO3 contains x g of NaOH and y g ...
Mass of NaOH = x
Moles of NaOH = x/40
Mass of Na2CO3 = y
Moles of Na2CO3 = y/106
x/40 = y/106
x + y = 4
x = 1.1, y = 2.9
x = 1.1 ≈ 1 (nearest integer)
Moles of NaOH = x/40
Mass of Na2CO3 = y
Moles of Na2CO3 = y/106
x/40 = y/106
x + y = 4
x = 1.1, y = 2.9
x = 1.1 ≈ 1 (nearest integer)
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4 g equimolar mixture of NaOH and Na2CO3 contains x g of NaOH and y g of Na2CO3. The value of x is _________ g. (Nearest integer)Correct answer is '1'. Can you explain this answer?
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