Question

A solution containing 15.60g of ethanol in 108.0 g of H₂O is prepared. Calculate the following Volumen .200L Explain the calculations

molar mass
volume of water
mass ethanol fraction
molarity of the solution
solution molality
molar fraction of water

molecular formula mass molar mass moles

solute
solvent
solution

Problema #4: (20) Se prepara una solución que contiene 15.60 g de etanol en 108.0 g de siguiente: etanol en 108.0 g de H20. Calcule lo Masa molar de etanol: g/mol Volumen de agua: volumen de la schein 2002 Fracción másica de etanol: Molaridad de la solución: Molalidad de la solución: Fracción molar de agua: Masa moles Fórmula Molecular Molar Masa (g/mol) Soluto Solvente H20 Solución

Answer 1

Given Mass of Salute Cethanal) - 15.60g. Mas Solvent (Water) = 118.DOG Total mass of Salution - 15.60+108.00 = 123.60 g ( 123.6 mL of solution density of Water mass Volume 108mL 1 Volume = of Water mass = 108 density I (2) Molar mass of water (120)=2+16=18 g/mol (3) Molar mass of Ethanol (CH ₂ OH) = 12 x 2 + 6X1 +16 = 469 | mol (4) Malarity of Solution - weight of salute & 1000 Molar mass of salute Volume of Salution in mL 46 123.6 5685.6. Molarity 2.74 mol/L 6 Molality of Solution Weight of Salute tona - Molar mass weight of Sol 27 Selena

(6) Mass traction of = Ethanol 15.6 x 100% 15.6+ 108 = 12.62 % D Mass traction of = 108 x 200% Water 123.6 = 87.38% (8 Molar fraction of Ethanal ME E nt no Male of Ethanol = n = We = 15.6 = 0.339 mole. ne 46 Mole of Water = new = Ww = 108 = 6.000 mol | HCMA, 0.39 - 0,399 - 0.05 35 0.339+6.000 6.339 100 Molar traction X = X-XE of Water (Mole fraction of solution X = 1) Then, xw = (1-0.053) xw = 0.9465
Density of water is 1g/ml,(volume of solution considered as equal to mass of solution).