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torisob [31]
3 years ago
12

a solution of unknown molecular substance is prepared by dissolving 0.50g of the unknown in 8.0g of benzene. the solution freeze

s at 3.9. determine the molar mass of the unknown
Chemistry
1 answer:
spayn [35]3 years ago
6 0

Answer:

THE MOLAR MASS OF THE UNKNOWN MOLECULAR SUBSTANCE IS 200 G/MOL.

Explanation:

Mass of the unknown substance = 0.50 g

Freezing point of the solution = 3.9 °C

Freezing point of pure benzene = 5.5 °C

Freezing point dissociation constant Kf = 5.12°C/m

First, calculate the temperature difference between the freezing point of pure benzene and the final solution freezing point.

Change in temperature = 5.5 -3.9 = 1.6 °C

Next is to calculate the number of moles or molarity of the compound that dissolved.

Using the formula:

Δt = i Kf m

Assume i = 1

So,

1.6 °C = 1 * 5.12 * x/ 0.005 kg of benzene

x = 1.6 * 0.008 / 5.12

x = 0.0128 / 5.12

x = 0.0025 moles.

Next is to calculate the molar mass using the formula, molarity = mass / molar mass

Molar mass = mass / molarity

Molar mass = 0.50 g /0.0025

Molar mass = 200 g/mol

Hence, the molar mass of the unknown compound is 200 g/mol

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For the galvanic (voltaic) cell Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s) (E°= 0.77 V at 25°C), what is [Fe2+] if [Mn2+] = 0.040 M and
avanturin [10]

Answer:

0.01836 M

Explanation:

Again the reaction equation is;

Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)

E°cell= 0.77 V

Ecell= 0.78 V

[Mn2+] = 0.040 M

[Fe2+] = the unknown

n=2

From Nernst's equation;

Ecell= E°cell- 0.0592/n log Q

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0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]

0.01/ -0.0296= log [Fe2+] /[0.040]

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0.459= [Fe2+] /[0.040]

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7 0
3 years ago
Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange
SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)

A

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate  = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

=\frac{5.75 g}{10.0 g}\times 100=57.5\%

57.5% was the percent yield.

C

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

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Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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