JEE MAIN - Chemistry (2023 - 11th April Evening Shift - No. 9)
A solution is prepared by adding $$2 \mathrm{~g}$$ of "$$\mathrm{X}$$" to 1 mole of water. Mass percent of "$$\mathrm{X}$$" in the solution is :
20%
10%
2%
5%
Explanation
First, we need to calculate the mass of 1 mole of water (H2O). The molecular weight of water is 18 g/mol, so 1 mole of water weighs 18 g.
The total mass of the solution is the mass of "X" plus the mass of water, which is 2 g + 18 g = 20 g.
We can calculate the mass percent of "X" in the solution with the following formula:
$$\text{Mass percent of "X"} = \left(\frac{\text{mass of "X"}}{\text{total mass of solution}}\right) \times 100\%$$
Substitute the given values into the formula :
$$\text{Mass percent of "X"} = \left(\frac{2 \, \text{g}}{20 \, \text{g}}\right) \times 100\% = 10\%$$
Therefore, Option B, $10\%$, is the correct answer.
The total mass of the solution is the mass of "X" plus the mass of water, which is 2 g + 18 g = 20 g.
We can calculate the mass percent of "X" in the solution with the following formula:
$$\text{Mass percent of "X"} = \left(\frac{\text{mass of "X"}}{\text{total mass of solution}}\right) \times 100\%$$
Substitute the given values into the formula :
$$\text{Mass percent of "X"} = \left(\frac{2 \, \text{g}}{20 \, \text{g}}\right) \times 100\% = 10\%$$
Therefore, Option B, $10\%$, is the correct answer.
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