What is the freezing point of a solution of 0.500 mol of sucros in 500.0 mL of water?

Answer: 0.245 moles of oxygen must be placed in the container to exert the given pressure at the given temperature. The Ideal Gas Law equation gives the relationship among the pressure, volume, temperature, and moles of gas.

Further Explanation:

The Ideal Gas Equation is:

$PV = nRT$

where:

P - pressure (in atm)

V - volume (in L)

n - amount of gas (in moles)

R - universal gas constant $0.08206 \frac{L-atm}{mol-K}$

T - temperature (in K)

In the problem, we are given the values:

P = 2.00 atm (3 significant figures)

V = 3.00 L (3 significant figures)

n = ?

T = 25.0 degrees Celsius (3 significant figures)

We need to convert the temperature to Kelvin before we can use the Ideal Gas Equation. The formula to convert from degree Celsius to Kelvin is:

$Temperature \ in \ Kelvin = Temperature\ in \ Celsius \ + \ 273.15$

Therefore, for this problem,

$Temperature\ in \ K = 25.0 +273.15\\Temperature\ in \ K = 298.15$

Solving for n using the Ideal Gas Equation:

$n \ = \frac{PV}{RT}\\n \ = \frac{(2.00 \ atm) \ (3.00 \ L)}{(0.08206 \ \frac{L-atm}{mol-K})( 298.15 \ K)} \\n \ = 0.245 \ mol$

The least number of significant figures is 3, therefore, the final answer must have only 3 significant figures.

Learn More

1. Learn more about Boyle's Law brainly.com/question/1437490

2. Learn more about Charles' Law brainly.com/question/1421697

3. Learn more about Gay-Lussac's Law brainly.com/question/6534668

Keywords: Ideal Gas Law, Volume, Pressure

4 0

3 years ago

Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -2

Mkey [24]

Answer:

ΔG°rxn = +50.8 kJ/mol

Explanation:

It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

ΔG°rxn = ΔH°rxn - T×S°rxn (1)

In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}

ΔH°rxn = 136.5kJ/mol

And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)

ΔH°rxn = 0.2875kJ/molK

And replacing in (1) at 298K:

ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK

ΔG°rxn = +50.8 kJ/mol

7 0

3 years ago

Read 2 more answers

How many moles of sodium ions are present in 0.25 kg of a 2.0 m solution of sodium chloride in water

LenKa [72]

0.50 is the answer on edge

8 0

4 years ago