Question

A solution of sucrose (molicular mass 342 u ) is prepared by dissolving 6.84 g in 100 g of water at 298 k respectively
1. calculate the boiling point of the solution.
Calculate freezing point of the solution

Answer 1

$\huge\bold{HEY:)-}$

Answer 2

The boiling point of the solution is $100.104^0C$ and freezing point of the solution is $-0.372^0C$

Explanation:

Elevation in boiling point

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

$\Delta T_b=T_b-T_b^0=(T_b-100)^0C$ = Elevation in boiling point

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

$K_b$ = boiling point constant = $0.52^0C/m$

m= molality

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

Weight of solvent (water)= 100 g = 0.1 kg

Molar mass of solute (sucrose) = 342 g/mol

Mass of solute (sucrose) = 6.84 g

$(T_b-100)^0C=1\times 0.52\times \frac{6.84g}{342g/mol\times 0.1kg}$

$(T_b-100)^0C=0.104$

$T_b=100.104^0C$

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(0-T_f)^0C$ = Depression in freezing point

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

$K_f$ = freezing point constant = $1.86^0C/m$

m= molality

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

Weight of solvent (water)= 100 g = 0.1 kg

Molar mass of solute (sucrose) = 342 g/mol

Mass of solute (sucrose) = 6.84 g

$(0-T_f)^0C=1\times 1.86\times \frac{6.84g}{342g/mol\times 0.1kg}$

$(0-T_f)^0C=0.372$

$T_f=-0.372^0C$

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