Answer:
Mole fraction: 0,0157.
Molality: 0,889m
Mass%: 16%
Explanation:
<em>The units are mole fraction, molality and mass%</em>
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Units of mole fraction are moles of glucose per total moles.
Moles of glucose assuming 1L are:
0,944 moles.
Moles of water in 1L are:
1L × (1,0624kg/L) × (1000g / 1kg) × (1mol / 18,02g) = <em>59,0 moles of water</em>
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Mole fraction is: 0,944 moles / (59,0 mol + 0,944mol) = <em>0,0157</em>
Molality is mole of solute (0,944) per kg of solution (1,0624kg):
0,944mol / 1,0624kg = <em>0,889m</em>
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In mass percent total mass is 1062,4g and mass of 0,944 moles of glucose is:
0,944mol×(180,156g/1mol) = 170g of glucose. Mass%:
170g / 1062,4g ×100 = <em>16%</em>
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I hope it helps!
