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vagabundo [1.1K]
3 years ago
5

The normal freezing point of cyclohexane is 6.55 C. When 0.458 g of benzophenone is dissolved in 15.0 g of cyclohexane, the free

zing point is found to be 3.19 C. What is the experimental molar mass of benzophenone? (Kf cyclohexane = 20.0 C m^-1).
A) 182
B) 18.0
C) 89.3
D) 160
E) 360
Chemistry
1 answer:
ella [17]3 years ago
3 0

Answer:

The experimental molar mass of benzophenone is 182 g/m (option A)

Explanation:

ΔT = Kf . m . i

(i means the Van't Hoff factor,  since benzophenone is not electrolyte, the value for i is 1)

ΔT = T° - T' (Melting temp sv pure - Melting temp sv in sl)

Kf is data (the cryoscopic constant)

m is molality (moles from solute in 1kg of solvent)

6,55°C - 3,19°C = 20,0° C/m . m

3,36 °C = 20,0 C/m . m

3,36 °C /20,0 m/C = m

0,168 = m

We have 0,168 moles of benzophenone in 1 kg of cyclohexane but we don't have 1kg, we have just 15 g so, we need the rule of three.

1000 g ______ 0,168 moles

15 g ________  (15 g . 0,168 m) / 1000 g = 0,00252 m

Now we can get the molar mass:

0,458 g of benzophenone / 0,00252 moles = 181,7 g/m

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Answer:

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2 years ago
If the mixing ratio of a parcel of air is 6 grams per kilogram and the relative humidity is 25 percent, the saturation mixing ra
Y_Kistochka [10]

Answer: -

24 grams per kilogram.

Explanation: -

We know that

The mixing ratio = actual (measured) mass of water vapor (in parcel) in grams / mass of dry (non water vapor) air (in parcel) in kilogram

The saturation mixing ratio = mass of water vapor required for saturation (in parcel) in grams/ mass of dry (non water vapor) air (in parcel) in kilograms

Relative humidity = actual (measured) water vapor content/ maximum possible water vapor amount (saturation)

Thus saturation mixing ratio = Mixing ratio / relative humidity

= 6 / (25/100)

= 24

4 0
3 years ago
What greek letter are used in the precise definition of a limit?
Ivan
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5 0
3 years ago
The Ksp of yttrium iodate, Y(IO3)3 , is 1.12×10−10 . Calculate the molar solubility, s , of this compound.
torisob [31]

Answer:

1.427x10^-3mol per L

Explanation:

Y(IO_{3} )_{3} ---- Y^{3+} +IO_{3} ^{3-}

I could use ⇌ in the math editor so I used ----

from the question each mole of Y(IO3)3 is dissolved  and this is giving us a mole of Y3+ and a mole of IO3^3-

Ksp = [Y^3+][IO3-]^3

So that,

1.12x10^-10 = [S][3S]^3

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1.12x10^-10 = 27S^4

the value of s is 0.001427mol per L

= 1.427x10^-3mol per L

so in conclusion

the molar solubility is therefore 1.427x10^-3mol per L

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Answer:

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Explanation:

If you draw the structure, you can see that there are two methyl groups and in between there.

Adjacent to CH3, there are four neighbouring hydrogens, therefore, n=4+1 = 5. The same is for methyl on other side. For carbon present in benzene ring,  there is 2, since there is one hydrogen on benzene per carbon.

4 0
3 years ago
Read 2 more answers
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