Example 2.1 Calculate the mole fraction of ethylene glycol (C_{2}*H_{6}*O_{2}) in a solution containing 20% of C_{2}*H_{6}*O_{2} by mass. Solution Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20 g of ethylene glycol and 80 g of water Molar mass of C_{2}*H_{6}*O_{2} = 12 * 2 + 1 * 6 + 16 * 2 = 62gmo * l ^ - 1 Moles of C_{2}*H_{6}*O_{2} = (20g)/(62gmo * l ^ - 1) = 0.322mol Molesofwater = (80g)/(18mo * l ^ - 1) = 4.44mol x elycal = moles of C 2 H 6 O 2 moles of C 2 H 0 O 2 +moles of H 2 O = (0.322mol)/(0.322mol + 4.444mol) = 0.068 Similarly, x mater = 4.444 mol 0.322 mol+4.444 mol =0.932 Mole fraction of water can also be calculated as: 1 - 0.068 = 0.932
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Example 2.1
Calculate the mole fraction of ethylene glycol (C_{2}*H_{6}*O_{2}) in a solution containing 20% of C_{2}*H_{6}*O_{2} by mass.
Solution Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20 g of ethylene glycol and 80 g of water Molar mass of C_{2}*H_{6}*O_{2} = 12 * 2 + 1 * 6 + 16 * 2 = 62gmo * l ^ - 1
Moles of C_{2}*H_{6}*O_{2} = (20g)/(62gmo * l ^ - 1) = 0.322mol
Molesofwater = (80g)/(18mo * l ^ - 1) = 4.44mol
x elycal = moles of C 2 H 6 O 2 moles of C 2 H 0 O 2 +moles of H 2 O
= (0.322mol)/(0.322mol + 4.444mol) = 0.068
Similarly, x mater = 4.444 mol 0.322 mol+4.444 mol =0.932
Mole fraction of water can also be calculated as: 1 - 0.068 = 0.932
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