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mariarad [96]
3 years ago
15

A sample of nitrogen occupies 10.0 liters at 25°C what would be the new volume at 20°C? 7.9 L Ob 9.8 L 10.2 L 10.6 L

Chemistry
1 answer:
PtichkaEL [24]3 years ago
7 0

Answer:

  9.4 liter

Explanation:

1) Data:

V₁ = 10.0 L

T₁ = 25°C = 25 + 273.15 K = 298.15 K

 P₁ = 98.7 Kpa

 T₂ = 20°C = 20 + 273.15 K = 293.15 K

  P₂ = 102.7 KPa

  V₂ = ?

2) Formula:

Used combined law of gases:

  PV / T = constant

  P₁V₁ / T₁ = P₂V₂ / T₂

3) Solution:

Solve the equation for V₂:

  V₂ = P₁V₁ T₂ / (P₂ T₁)

Substitute and compuite:

V₂ = P₁V₁ T₂ / (P₂ T₁)

V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)

V₂ =  9.4 liter ← answer

You can learn more about gas law problems reading this other answer on

Explanation:

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<u>Answer:</u> The vapor pressure of solution is 459.17 mmHg

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

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Given mass of testosterone = 7.752 g

Molar mass of testosterone = 288.4 g/mol

Putting values in equation 1, we get:

\text{Moles of testosterone}=\frac{7.752g}{288.4g/mol}=0.027mol

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Given mass of diethyl ether = 208.0 g

Molar mass of diethyl ether = 74.12 g/mol

Putting values in equation 1, we get:

\text{Moles of diethyl ether}=\frac{208.0g}{74.12g/mol}=2.81mol

Mole fraction of a substance is calculated by using the equation:

\chi_A=\frac{n_A}{n_A+n_B}

\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}

\chi_{\text{testosterone}}=\frac{0.027}{0.027+2.81}\\\\\chi_{\text{testosterone}}=0.0095

The formula for relative lowering of vapor pressure will be:

\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}

where,

p^o = vapor pressure of solvent (diethyl ether) = 463.57 mmHg

p^s = vapor pressure of the solution = ?

i = Van't Hoff factor = 1 (for non electrolytes)

\chi_{\text{solute}} = mole fraction of solute (testosterone) = 0.0095

Putting values in above equation, we get:

\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg

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