**Calculate the amount in moles of a gas which occupies 10.5 dm**^{3} at 6 atm and 30^{o}C...

^{3}at 6 atm and 30

^{o}C...

### Question

Calculate the amount in moles of a gas which occupies 10.5 dm^{3}at 6 atm and 30

^{o}C [R = 082 atm dm

^{3}K

^{-1}mol

^{-1}]

### Options

A) 2.536

B) 1.623

C) 4.736

D) 0.394

Related Lesson: Calculating Moles and Mass | Quantitative Aspects of Chemical Change

The correct answer is A.

### Explanation:

For an ideal gas PV = nRT

Amount in moles = n

Volume v = 10.5dm^{3}

Pressure P = 6atm

Temperature T = 30°C + 273 = 303k

R, Gas constant = 0.082 atmdm^{3}k^{-1} mol

Recall from ideal gas equation

pv = nRT

n = \(\frac{RV}{RT}\)

n = \(\frac{6 \times 105}{0.082 \times 303}\)

n= 2.536mol

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## Discussion (2)

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Pressure [P]= 6 atm

Volume [V]= \( 10.5dm^3 \)

n= number of moles

Gas Constant [R]= \( 0.082atm dm^{3}K^{-1}mol^{-1} \)

Temperature [T]=30°c=30+273K=303K

Using the Ideal Gas equation;

PV=nRT

n=\( \frac{PV}{RT} \)

n= \( \frac{6× 10.5}{0.082×303} \)

n=\( \frac{63}{24.846} \)

n=2.5356moles

n≈2.536 moles

Option [A] is 💯% correct.

For an ideal gas PV = nRT

Amount in moles = n

Volume v = 10.5dm

^{3}Pressure P = 6atm

Temperature T = 30°C + 273 = 303k

R, Gas constant = 0.082 atmdm

^{3}k^{-1}molRecall from ideal gas equation

pv = nRT

n = \(\frac{RV}{RT}\)

n = \(\frac{6 \times 105}{0.082 \times 303}\)

n= 2.536mol