1) A solution is made by dissolving 6.35g of glucose (C6H1206) in 25.0 mL of wat

Question

1) A solution is made by dissolving 6.35g of glucose (C6H1206) in 25.0 mL of water. The density of water is 1.00 g/mL. Calculate the molality of the solution

2) A solution contains 5.0g of toluene (C7H8) and 255g of benzene and has a density of 0.876 g/mL. Calculate the molarity of the solution

3) A solution is made containing 7.5g CH3OH in 275g H20. Assume the density of water as 1.00 g/mL

A) calculate the mole fraction of CH3OH

B) The mass percent of CH3OH

C) The molality of CH3OH

Thank you for whoever is able to help!

Explanation / Answer

(1) Moles of glucose = mass/molar mass of glucose

= 6.35/180.16 = 0.035246 mol


Mass of water = volume x density of water

= 25.0 x 1.00 = 25.0 g = 0.0250 kg


Molality = moles of glucose/mass of water in kg

= 0.035246/0.0250

= 1.41 m


(2) Moles of toluene = mass/molar mass of toluene

= 5.0/92.14 = 0.054265 mol


VOlume of solution = mass/density of solution

= (5.0 + 255)/0.876 = 296.8 mL = 0.2968 L


Molarity = moles of toluene/volume of solution

= 0.054265/0.2968

= 0.183 M (approximately 0.18 M)


(3) (A) Moles of CH3COH = mass/molar mass of CH3OH

= 7.5/32.04 = 0.23408 mol


Moles of H2O = mass/molar mass of H2O

= 275/18.02 = 15.26082 mol


Mole fraction = moles of CH3OH/total moles

= 0.23408/(0.23408 + 15.26082)

= 0.0151 (approximately 0.015)


(B) Mass percent = mass of CH3OH/total mass x 100%

= 7.5/(7.5 + 275) x 100%

= 2.65% (approximately 2.7%)


(C) Molality = moles of CH3OH/mass of H2O in kg

= 0.23408/0.275

= 0.851 m (approximately 0.85 m)

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