1. What is the molarity of a solution of 50.0 g of propanol (CH 3 CH 2 CH 2 OH)

Question

1. What is the molarity of a solution of 50.0 g of propanol (CH3CH2CH2OH) in 152 mL water, if the density of water is 1.00 g/mL?  

a. 5.47 M

b. 0.00547 M

c. 0.833 M

d. 0.183 M

e. none of these

2. What is the mole percent of ethanol (C2H5OH), which consists of 71.0 g of ethanol for every 10.0 g of water present?

a. 71.0%

b. 73.5%

c. 26.5%

d. 22.1%

e. 87.7%

3.  Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is a renewable resource and ethanol-blended fuels, like gasohol, appear to burn more efficiently in combustion engines. The heat of combustion of ethanol is 326.7 kcal/mol.

The heat of combustion of heptane, C7H16, is 1.151×103 kcal/mol. How much energy is released during the complete combustion of 348 grams of heptane?

Assuming the same efficiency, would 348 grams of ethanol provide more, less, or the same amount of energy as 348 grams of heptane? More, less, or the same amount? Please type the answers! :)

Explanation / Answer

1)

Molar mass of CH3CH2CH2OH = 3*MM(C) + 8*MM(H) + 1*MM(O)

= 3*12.01 + 8*1.008 + 1*16.0

= 60.094 g/mol

mass of CH3CH2CH2OH = 50.0 g

we have below equation to be used:

number of mol of CH3CH2CH2OH,

n = mass of CH3CH2CH2OH/molar mass of CH3CH2CH2OH

=(50.0 g)/(60.094 g/mol)

= 0.832 mol

volume , V = 152 mL

= 0.152 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 0.832/0.152

= 5.47 M

Answer: a

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