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.

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

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

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

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

m= molality

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

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

$16.6^0C=1\times 1.22\times \frac{x}{18g/mol\times 0.4734kg}$

$x=116g$

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.