JEE MAIN - Chemistry (2023 - 13th April Evening Shift - No. 14)
$$0.400 \mathrm{~g}$$ of an organic compound $$(\mathrm{X})$$ gave $$0.376 \mathrm{~g}$$ of $$\mathrm{AgBr}$$ in Carius method for estimation of bromine. $$\%$$ of bromine in the compound $$(\mathrm{X})$$ is ___________.
(Given: Molar mass $$\mathrm{AgBr=188~g~mol^{-1}}$$
$$\mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}$$)
Answer
40
Explanation
To calculate the percentage of bromine in the compound X, we first need to determine the moles of AgBr formed and then the moles of bromine present in the compound. Finally, we can find the percentage of bromine by weight.
1. Calculate the moles of AgBr formed:
Moles of AgBr = mass / molar mass = 0.376 g / 188 g/mol = 0.002 mol
Since 1 mol of AgBr contains 1 mol of Br, the moles of Br in the compound X are also 0.002 mol.
2. Calculate the mass of bromine in the compound:
Mass of Br = moles of Br × molar mass of Br = 0.002 mol × 80 g/mol = 0.16 g
3. Calculate the percentage of bromine in the compound X:
Percentage of Br = (mass of Br / mass of compound X) × 100 = (0.16 g / 0.400 g) × 100 = 40%
So, the percentage of bromine in the compound X is 40%.
1. Calculate the moles of AgBr formed:
Moles of AgBr = mass / molar mass = 0.376 g / 188 g/mol = 0.002 mol
Since 1 mol of AgBr contains 1 mol of Br, the moles of Br in the compound X are also 0.002 mol.
2. Calculate the mass of bromine in the compound:
Mass of Br = moles of Br × molar mass of Br = 0.002 mol × 80 g/mol = 0.16 g
3. Calculate the percentage of bromine in the compound X:
Percentage of Br = (mass of Br / mass of compound X) × 100 = (0.16 g / 0.400 g) × 100 = 40%
So, the percentage of bromine in the compound X is 40%.
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