Calculatethemassof 3-011-1023Natoms-

Numberofmoles(n)=Givennumberofparticles(N)/Avogadro'snumber(N_A)0ex0exn=3.011×10^23atoms/6.022×10^23atomsmol^ 10ex0exn=0.5 mole0ex0exm=n×M0ex0exm=0.5 mol× 14 g/ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Nucleus: Common Terms

Calculatethem...

Question

$Calculate the mass of 3.011 \times 10^{23} N atoms .$

Open in App

Solution

$Number of moles (n) = \frac{Given number of particles (N)}{Avogadro' s number (N_{A})} n = \frac{3.011 \times 10^{23} atoms}{6.022 \times 10^{23} atoms {mol}^{- 1}} n = 0.5 mole m = n \times M m = 0.5 mol \times 14 g / mol m = 7 g Therefore, mass of 3.011 \times 10^{23} N atoms is 7 g .$

Suggest Corrections

Similar questions

Consider a $70 %$ efficient hydrogen-oxygen cell working under the standard condition at $1$ bar and $298$ K. Its cell reaction is

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$

The work derived from the cell on the consumption of $1.0 \times 10^{- 3}$ $mol of H_{2} (g)$ is used to compress $1.00 mol$ of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in $K$ ) of the ideal gas?

The standard reduction potentials for the two half-cells are given below

$O_{2} (g) + 4 H^{+} (aq) + 4 e^{-} \to 2 H_{2} O (l)$ , $E^{0}$ = $1.23 V$

$2 H^{+} (aq) + 2 e^{-} \to H_{2} (g)$ , $E^{0}$ = $0.00 V$

Use $F = 96500 {Cmol}^{- 1}$ , $R = 8.314 {Jmol}^{- 1} K^{- 1}$ .

Join BYJU'S Learning Program

Calculatethemassof3.011×1023Natoms.

Numberofmoles(n)=Givennumberofparticles(N)Avogadro'snumber(NA)n=3.011×1023atoms6.022×1023atomsmol-1n=0.5molem=n×Mm=0.5mol×14g/molm=7gTherefore,massof3.011×1023Natomsis7g.

$Calculate the mass of 3.011 \times 10^{23} N atoms .$

$Number of moles (n) = \frac{Given number of particles (N)}{Avogadro' s number (N_{A})} n = \frac{3.011 \times 10^{23} atoms}{6.022 \times 10^{23} atoms {mol}^{- 1}} n = 0.5 mole m = n \times M m = 0.5 mol \times 14 g / mol m = 7 g Therefore, mass of 3.011 \times 10^{23} N atoms is 7 g .$