Question

What volume of water (solute) (d = 1.00 g/mL) should be added to 600. mL of ethanol (solvent, C2H5OH) in order to have a solution that boils at 95.0°C? [For ethanol, Kb = 1.22 °C/m, density = 0.789 g/ mL, boiling point = 78.4°C]

Answer 1

Answer: Volume of water to be added is 116 ml.

Step-by-step explanation:

Elevation in boiling point is given by:

`\Delta T_b=i* K_b* m`

`\Delta T_b=T_b-T_b^0=(95.0-78.4)^0C=16.6^0C` = elevation in boiling point

i= vant hoff factor = 1 (for non electrolyte)

`K_b` =boiling point constant =
`1.22^0C/m`

m= molality

`\Delta T_b=i* K_b* \frac{\text{mass of solute}}{\text{molar mass of solute}}* \text{weight of solvent in kg}}`

Density of solvent =
`\frac{\text {mass of solvent}}{\text {Volume of solvent}}`

`0.789g/ml=\frac{\text {mass of solvent(ethanol)}}{600ml}`

`{\text {mass of solvent(ethanol)}=473.4g=0.4734 kg` (1kg=1000g)

Mass of solute (water) = x g

`16.6^0C=1* 1.22* (x)/(18g/mol* 0.4734kg)`

`x=116g`

Density of solute =
`\frac{\text {mass of solute}}{\text {Volume of solute}}`

`1.0g/ml=\frac{116g}{\text {Volume of solute(water)}}`

`{\text {Volume of solute(water)}}=116ml`

Thus the volume of water to be added is 116 ml.