Question

Calculate the number of moles & molecules of ammonia present in 5.6 dm3 of its volume at STP.​

Answer 1

$\mathfrak{\huge{\underline{Answer:}}}$

Explanation:

THE NUMBER OF MOLES AND MOLECULES OF AMMONIA AT STP:

Given :V = 5.6 dm3 = 5.6L

Then ,We know that

n = V / 22.4L

n= 5.6 / 22.4

n = 56 / 224

n = 1 / 4

n = 0.25

No. Of molecules = n × avagadro number

=0.25*6.022*10^²³

=1.5*10^²³

Answer 2

The number of moles and molecules present in 5.6 $dm^{3}$ of ammonia are 0.25 moles and $1.505X10^{23}$ molecules respectively.

Given,

Volume of ammonia=5.6 $dm^{3}$.

To find,

the number of moles & molecules of ammonia present in 5.6 dm3 of its volume at STP.​

Solution:

• STP conditions represent the standard temperature and pressure where the temperature is equal to 273 K(Kelvin) and the pressure is equal to 1 atm(1 bar).
• 1 atm=1.01325 bar.
• Moles of a gas is equal to the ratio of volume of the gas at STP(in litres) and 22.4 litres.
• $Moles=\frac{Volume(L)}{22.4}$.
• 1 $dm^{3}$=1 L.
• 5.6 $dm^{3}$=5.6 L.

$Moles=\frac{Volume(L)}{22.4}\\Moles=\frac{5.6}{22.4}\\Moles=\frac{1}{4}\\Moles=0.25.$

0.25 moles of ammonia are present.

1 mole of a substance contains Na number of molecules.

$Na=6.02X10^{23} .$

0.25 moles of ammonia will contain,

$=0.25XNa\\=0.25X6.02X10^{23} \\=1.505X10^{23}.$

5.6 $dm^{3}$ of ammonia contains 0.25 moles and $1.505X10^{23}$ molecules.

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