Calculate the energy required to convert all atoms in 4.8 g of Mg to M g2+ in the vapor state. I E1 and I E2 of M g are 740 kJ / mol and 1450 kJ / mol respectively.

Question

Calculate the energy required to convert all atoms in 4.8 g of Mg to M g2+ in the vapor state. I E1 and I E2 of M g are 740 kJ / mol and 1450 kJ / mol respectively. (A) +740 kJ / mol (B) -740 kJ / mol (C) -1450 kJ / mol (D) +438 kJ / mol

Tardigrade Admin · Accepted Answer

For 1 mole of (M g arrow M g2+) I E=I E1+I E2=(740+1450)=2190 kJ / mol Number of mole in 4.8 g of M g=4.8 / 24=0.2 mol For 1 mole energy required =2190 kJ / mol For 0.2 mole energy required =2190 × 0.2=438 kJ Thus, for 4.8 g of M g to M g2+ conversion, energy required is 438 kJ.

Q. Calculate the energy required to convert all atoms in 4.8g of Mg to Mg2+ in the vapor state. IE1​ and IE2​ of Mg are 740kJ/mol and 1450kJ/mol respectively.

+740 kJ / mol

-740 kJ / mol

-1450 kJ / mol

+438 kJ / mol

For 1 mole of (Mg→Mg2+) IE=IE1​+IE2​=(740+1450)=2190kJ/mol Number of mole in 4.8g of Mg=4.8/24=0.2mol For 1 mole energy required =2190kJ/mol For 0.2 mole energy required =2190×0.2=438kJ Thus, for 4.8g of Mg to Mg2+ conversion, energy required is 438kJ.

Q. Calculate the energy required to convert all atoms in $4.8 g$ of $M g$ to $M g^{2 +}$ in the vapor state. $I E_{1}$ and $I E_{2}$ of $M g$ are $740 k J / m o l$ and $1450 k J / m o l$ respectively.