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Brrunno [24]
3 years ago
10

A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg

.mol-1. A solution is prepared by dissolving some urea ((NH2)2CO) in 950. g of X. This solution freezes at -13.6 C. calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.
Chemistry
1 answer:
Brut [27]3 years ago
8 0

Answer:  1.0\times 10^2g

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=(-6.4-(13.6))^0C=7.2^0C = Depression in freezing point

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

K_f = freezing point constant = 3.96^0C/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 (X)= 950 g = 0.95 kg  

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}

x=1.0\times 10^2g

Thus 1.0\times 10^2g urea was dissolved.

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