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zloy xaker [14]
3 years ago
14

Where does the Grevy's Zebra live?

Chemistry
2 answers:
Komok [63]3 years ago
7 0
Rainforest with lots of moisture and humidity is you answer
allsm [11]3 years ago
7 0
2. Rainforests with lots of moisture and humidity
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3 years ago
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A certain substance X has a normal freezing point of -10.1 degree C and a molal freezing point depression constant Kf = 5.32 °C.
Citrus2011 [14]

Answer : The freezing point of a solution is -15.4^oC

Explanation : Given,

Molal-freezing-point-depression constant (K_f) = 5.32^oC/m

Mass of urea (solute) = 29.82 g

Mass of solvent = 500 g  = 0.500 kg

Molar mass of urea = 60.06 g/mole

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of solvent in Kg}}

where,

\Delta T_f = change in freezing point

\Delta T_s = freezing point of solution = ?

\Delta T^o = freezing point of solvent = -10.1^oC

i = Van't Hoff factor = 1 (for urea non-electrolyte)

K_f = freezing point constant = 5.32^oC/m

m = molality

Now put all the given values in this formula, we get

-10.1^oC-T_s=1\times (5.32^oC/m)\times \frac{29.82g}{60.06g/mol\times 0.500kg}

T_s=-15.4^oC

Therefore, the freezing point of a solution is -15.4^oC

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