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marusya05 [52]
4 years ago
5

George is at a water park with his family. He needs to get a tube to use for one of the rides, and he chose his color first. The

re are black tubes, dark purple tubes, dark blue tubes, and yellow tubes. Each member of his family had to choose a different colored tube. George chose a yellow tube, because he thought it would be the coolest. After everyone chose a tube, he touched all of the tubes. The yellow tube was the coolest. How do you think George knew that the yellow tube would be the coolest?
A) He knew that dark colors absorb energy, so they would be coolest.
B) He knew that dark colors reflect energy, so they would be coolest.
C) He knew that light colors absorb energy, so they would be coolest.
D) He knew that light colors reflect energy, so they would be coolest.
Chemistry
1 answer:
Elza [17]4 years ago
3 0
The answer would be D
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Nesterboy [21]

The given question is incomplete. The complete question is:

When 282. g of glycine (C2H5NO2) are dissolved in 950. g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 282. g of ammonium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for ammonium chloride in X.

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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=8.2^0C = Depression in freezing point  

K_f = freezing point constant = ?

i = 1 ( for non electrolyte)

m= molality

8.2^0C=1\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 solute (glycine) = 75.07 g/mol

Mass of solute (glycine) = 282 g

8.2^0C=1\times K_f\times \frac{282g}{75.07g/mol\times 0.95kg}

K_f=2.07

ii) 20.0^0C=i\times \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 solute (ammonium chloride) = 53.49 g/mol

Mass of solute (ammonium chloride) = 282 g

20.0^0C=i\times 2.07\times \frac{282g}{53.49g/mol\times 0.95kg}

i=1.74

Thus the van't Hoff factor for ammonium chloride is 1.74

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