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adelina 88 [10]
3 years ago
7

If the volume and moles of a container are held constant, the temperature of a gas is inversely proportional to the pressure.

Chemistry
1 answer:
noname [10]3 years ago
4 0

Answer:

FALSE

Explanation:

Assuming that the gas is ideal

Therefore the gas obeys the ideal gas equation

<h3>Ideal gas equation is </h3><h3>P × V = n × R × T</h3>

where

P is the pressure exerted by the gas

V is the volume occupied by the gas

n is the number of moles of the gas

R is the ideal gas constant

T is the temperature of the gas

Here volume of the gas will be the volume of the container

Given the volume of the container and number of moles of the gas are constant

As R will also be constant, the pressure of the gas will be directly proportional to the temperature of the gas

P ∝ T

∴ Pressure will be directly proportional to the temperature

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A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol
katrin2010 [14]

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of 0.80^oC and a freezing point depression constant K_f=7.82^oC.kg/mol . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

<u>Answer:</u> The freezing point of the solution is -17.6^oC

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m

OR

\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

where,

Freezing point of pure solvent = 0.80^oC

Freezing point of solution = ?^oC

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

K_f = freezing point depression constant = 7.82^oC/m

m_{solute} = Given mass of solute (iron (III) chloride) = 81.1 g

M_{solute} = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

w_{solvent} = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC

Hence, the freezing point of the solution is -17.6^oC

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