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We know that Molarity = No . of moles Volume of.pdf

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We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L ---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- & 2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5 mol Solution We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L ---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- & 2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5 mol.

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We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L ---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- & 2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5 mol Solution We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L ---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- & 2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5 mol

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  • 1. We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L ---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- & 2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5 mol Solution We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L ---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- & 2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5 mol