JEE MAIN - Chemistry (2021 - 20th July Evening Shift - No. 20)
4g equimolar mixture of NaOH and Na2CO3 contains x g of NaOH and y g of Na2CO3. The value of x is ____________ g. (Nearest integer)
Answer
1
Explanation
Mass of NaOH = x
Moles of NaOH = $${x \over {40}}$$
Mass of Na2CO3 = y
Moles of Na2CO3 = $${y \over {106}}$$
$${x \over {40}} = {y \over {106}}$$
x + y = 4
x = 1.1, y = 2.9
x = 1.1 $$ \approx $$ 1 (nearest integer)
Moles of NaOH = $${x \over {40}}$$
Mass of Na2CO3 = y
Moles of Na2CO3 = $${y \over {106}}$$
$${x \over {40}} = {y \over {106}}$$
x + y = 4
x = 1.1, y = 2.9
x = 1.1 $$ \approx $$ 1 (nearest integer)
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