a sugar syrup of weight 214.2g contain 34.2g of sugar calculate molality and mole fraction
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(i) Molal concentration = Molality
= Moles of solute/Mass of solvent in Kg
(ii) Mole fraction of sugar
= Moles of sugar/Total moles in solution
(i) Weight of sugar syrup = 214.2 g
Weight of sugar in the syrup = 34.2 g
∴ Weight of water in the syrup = 214.2 – 34.2 = 180.0 g
Mol. wt. of sugar, C12H22O11 = 342
∴Molal concentration = 34.2 *1000/342 *180 = 0.56
(ii) Mol. wt. water, H2O = 18
∴ Mole fraction of sugar = 34.2/342 /180/18+34.2/342
= 0.1/10+0.1 = 0.1/10.1 = 0.0099
= Moles of solute/Mass of solvent in Kg
(ii) Mole fraction of sugar
= Moles of sugar/Total moles in solution
(i) Weight of sugar syrup = 214.2 g
Weight of sugar in the syrup = 34.2 g
∴ Weight of water in the syrup = 214.2 – 34.2 = 180.0 g
Mol. wt. of sugar, C12H22O11 = 342
∴Molal concentration = 34.2 *1000/342 *180 = 0.56
(ii) Mol. wt. water, H2O = 18
∴ Mole fraction of sugar = 34.2/342 /180/18+34.2/342
= 0.1/10+0.1 = 0.1/10.1 = 0.0099
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Answer:
(i) Molal concentration = Molality
= Moles of solute/Mass of solvent in Kg
(ii) Mole fraction of sugar
= Moles of sugar/Total moles in solution
(i) Weight of sugar syrup = 214.2 g
Weight of sugar in the syrup = 34.2 g
∴ Weight of water in the syrup = 214.2 – 34.2 = 180.0 g
Mol. wt. of sugar, C12H22O11 = 342
∴Molal concentration = 34.2 *1000/342 *180 = 0.56
(ii) Mol. wt. water, H2O = 18
∴ Mole fraction of sugar = 34.2/342 /180/18+34.2/342
= 0.1/10+0.1 = 0.1/10.1 = 0.0099
Explanation:
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