JEE MAIN - Chemistry (2020 - 3rd September Morning Slot - No. 10)
The mole fraction of glucose (C6H12O6
) in an aqueous binary solution is 0.1. The mass percentage of
water in it, to the nearest integer, is _______.
Answer
47
Explanation
Mole fraction of glucose in aqueous solution
= 0.1
Let total mole is 1 mol then mole of glucose will be 0.1 and mole of water will be 0.9.
So mass % of water = $${{0.9 \times 18} \over {0.1 \times 180 + 0.9 \times 18}}$$ $$ \times $$ 100
= 47.37 $$ \simeq $$ 47
Let total mole is 1 mol then mole of glucose will be 0.1 and mole of water will be 0.9.
So mass % of water = $${{0.9 \times 18} \over {0.1 \times 180 + 0.9 \times 18}}$$ $$ \times $$ 100
= 47.37 $$ \simeq $$ 47
Comments (0)
