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Radda [10]
3 years ago
12

A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is

exactly 750.0 g. The boiling point of this solution is 100.220 ∘ C . What is the molecular weight of the unknown compound?
Chemistry
1 answer:
e-lub [12.9K]3 years ago
8 0

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

  • ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

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Need help !!!!! ASAP
Korolek [52]
<h2>Hello!</h2>

The answer is:

The new volume is 2.84 L.

V_{2}=2.84L

<h2>Why?</h2>

To solve the problem, we need to remember what STP means. STP means that the gas is at standard temperature and pressure, or 273.15 K (0°C) and 1 atm.

Also, we need to use the Combined Gas Law, since the temperature, the pressure and the volume are being changed.

The Combined Gas Law establishes a relationship between the temperature, the pressure and the volume of an ideal gas using , Gay-Lussac's Law, Charles's Law,  and Boyle's Law.

The law is defined by the following equation:

\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}

Where,

P_{1} is the first pressure.

V_{1} is the first volume.

T_{1} is the first temperature.

P_{2} is the second pressure.

V_{2} is the second volume.

T_{2} is the second temperature.

So, we are given the following information:

V_{1}=2L\\P_{1}=1atm\\T_{1}=0\°C=273.15K\\P_{2}=80kPa=0.8atm\\T_{2}=37\°C=37+273=310K

Then, isolating the new volume, and substituting, we have:

\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}\\\\V_{2}=\frac{P_{1}V{1}}{T_{1}}*\frac{T_{2}}{P_{2}}\\\\V_{2}=\frac{1atm*2L}{273.15K}*\frac{310K}{0.8kPa}=2.84L

Hence, the new volume is 2.84 L.

V_{2}=2.84L

Have a nice day!

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4 years ago
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