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worty [1.4K]
3 years ago
15

The density of benzene at 15 ∘C∘C is 0.8787 g/mL. Calculate the mass of 0.1700 LL of benzene at this temperature. Express your a

nswer using four significant figures.
Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
5 0

Answer:

Mass of benzene is: 149.3 g

Explanation:

Let's use density to calculate mass.

Density = Mass / Volume

Mass = Density . volume

Be careful, because density is in g/mL and the volume is in L. So let's convert the L to mL: 0.170 L . 1000 mL / 1L = 170 mL

0.8787 g/mL . 170 mL = Mass of benzene

Mass of benzene is: 149.3 g

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If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

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Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

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The volume of alcohol which is placed in a small laboratory = 1.0 L

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Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

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= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

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The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

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