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lara [203]
3 years ago
10

A sealed helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 293 K is moved. What volume will the bal

loon occupy at a place where the pressure is 0.600 atm and the temperature is 253 K?
Chemistry
1 answer:
bixtya [17]3 years ago
8 0

Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V =  Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

V' = PVT'/P'T......................... Equation 2

Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

Substitute these values into equation 2

V' = (4.5×1×253)/(0.6×293)

V' = 1138.5/175.8

V' = 6.48 L

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<u>Answer:</u> The molality of non-electrolyte is 24.69 m

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We are given:

Mole fraction of saturated aqueous solution = 0.310

This means that 0.310 moles of non-electrolyte is present.

Moles of water (solvent) = 1 - 0.310 = 0.690 moles

To calculate the mass from given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of water = 0.690 moles

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

0.690mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.690mol\times 18g/mol)=12.42g

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}

Where,

n_{solute} = Moles of solute (non-electrolyte) = 0.310 moles

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Putting values in above equation, we get:

\text{Molality of non-electrolyte}=\frac{0.310\times 1000}{12.42}\\\\\text{Molality of non-electrolyte}=24.96m

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