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A mixture is prepared by mixing 10 gm H(...

A mixture is prepared by mixing 10 gm `H_(2)SO_(4) and 40 gm SO_(3)` calculate mole fraction of `H_(2)SO_(4)`

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To calculate the mole fraction of \( H_2SO_4 \) in the mixture of \( H_2SO_4 \) and \( SO_3 \), we will follow these steps: ### Step 1: Calculate the number of moles of \( H_2SO_4 \) The formula to calculate the number of moles is: \[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \] For \( H_2SO_4 \): - Given mass = 10 g - Molar mass of \( H_2SO_4 = 2 + 32 + (16 \times 4) = 98 \, g/mol \) Now, we can calculate the moles: \[ \text{Moles of } H_2SO_4 = \frac{10 \, g}{98 \, g/mol} \approx 0.102 \, moles \] ### Step 2: Calculate the number of moles of \( SO_3 \) For \( SO_3 \): - Given mass = 40 g - Molar mass of \( SO_3 = 32 + (16 \times 3) = 80 \, g/mol \) Now, we can calculate the moles: \[ \text{Moles of } SO_3 = \frac{40 \, g}{80 \, g/mol} = 0.5 \, moles \] ### Step 3: Calculate the total number of moles in the mixture Now, we can find the total number of moles: \[ \text{Total moles} = \text{Moles of } H_2SO_4 + \text{Moles of } SO_3 \] \[ \text{Total moles} = 0.102 \, moles + 0.5 \, moles = 0.602 \, moles \] ### Step 4: Calculate the mole fraction of \( H_2SO_4 \) The mole fraction (\( \chi \)) is given by the formula: \[ \chi_{H_2SO_4} = \frac{\text{Moles of } H_2SO_4}{\text{Total moles}} \] Substituting the values we calculated: \[ \chi_{H_2SO_4} = \frac{0.102 \, moles}{0.602 \, moles} \approx 0.169 \] ### Final Answer The mole fraction of \( H_2SO_4 \) in the mixture is approximately **0.169**. ---
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The percentage labelling (mixture of H_(2)SO_(4) and SO_(3)) refers to the total mass of pure H_(2)SO_(4) . The total amount of H_(2)SO_(4) found after adding calculated amount of water to 100 g oleum is the percentage labelling of oleum.Higher the percentage lebeling of oleum higher is the amount of free SO_(3) in the oleum sample. 100 g sample of '149%' oleum was taken and calculated amount of H_(2)O was added to make H_(2)SO_(4) . 500 mL solution of x MKOH solution is required to neutralize the solution. The value of x is.

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ALLEN- CONCENTRATION TERMS-Exercise S - I
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