Question

In sulphur estimation, \(2.0 \times 10^{-3}\) mol of an organic compound (X) (molar mass 76 \(g \cdot mol^{-1}\)) gave 0.4813 g of barium sulphate (molar mass 233 \(g \cdot mol^{-1}\)). The percentage of sulphur in the compound (X) is _______ \(\times 10^{-1}\) % (Nearest integer)

Answer 1

Correct Answer: 435

Step 1: Find mass of compound $X$ used
Given: \[ n_X=2.0\times 10^{-3}\ \text{mol} \] \[ M_X=76\ \text{g mol}^{-1} \] Mass of compound: \[ m_X=n_X\times M_X \] \[ =(2.0\times 10^{-3})\times 76 \] \[ =0.152\ \text{g} \]
Step 2: Find moles of $BaSO_4$ formed
Given: \[ m_{BaSO_4}=0.4813\ \text{g} \] \[ M_{BaSO_4}=233\ \text{g mol}^{-1} \] \[ n_{BaSO_4} = \frac{m_{BaSO_4}}{M_{BaSO_4}} \] \[ = \frac{0.4813}{233} \] \[ =2.0657\times 10^{-3}\ \text{mol} \]
Step 3: Find mass of sulphur
Each mole of $BaSO_4$ contains one mole of sulphur. So, \[ n_S=n_{BaSO_4} = 2.0657\times 10^{-3}\ \text{mol} \] Mass of sulphur: \[ m_S=n_S\times 32 \] \[ = (2.0657\times 10^{-3})\times 32 \] \[ =0.0661\ \text{g} \]
Step 4: Calculate percentage of sulphur
\[ %S= \frac{m_S}{m_X}\times 100 \] \[ = \frac{0.0661}{0.152}\times 100 \] \[ =43.49% \]
Step 5: Write in required form
Question asks in the form \[ k\times 10^{-1}% \] So, \[ 43.49% = 434.9\times 10^{-1}% \] Nearest integer: \[ k=435 \] Final Answer: \[ \boxed{435} \]