Question

1. Molal depression constant of water is 1.86 K Kg me
0.02 mol of urea dissolved in 100 g of water
produce a depression in freezing point of:​

Answer 1

Answer:

0.372kelwin

0.02 mol urea in 100g of water.

multiplying with 10

0.2 mol in 1000g/1kg water

thus, molality=0.2mol/kg

depression in freezing point=molality×molal depression constant=0.2×1.86=0.372kelwin

Answer 2

0.02 mol of urea dissolved in 100 g of water  produce a depression in freezing point of 273.372 K

Explanation:

Depression in freezing point is given by:

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

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

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

$K_f$ = freezing point constant = $1.86K/m$

m= molality

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

moles of solute (urea) = 0.02

Molar mass of solute (urea)= 60 g/mol

Weight of solvent (water) = 100 g = 0.1 kg    (1kg=1000g)

$(T_f^0-T_f)=1\times 1.86\times \frac{0.02}{0.1kg}$

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

$T_f=0.372^0C=273.372K$      $0^0C=273K$

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