Question

9.7*10^17 atoms of fe weight as much as 1cm^3 og h2 what is the atomic mass

Answer 1

Answer:

The atomic mass of iron is 55.49 g/mol.

Explanation:

1 cc of $H_2=1cm^3=0.001 L$  

This is the volume of the hydrogen gas at STP.

At STP, Pressure,P = 1 atm

At STP, Temperature,T = 273 K

$PV=nRT$

$1 atm\times 0.001 L=n\times 0.0820 L atm/ mol K\times 273 K$

$n = 4.4670\times 10^{-5} mol$

Mass of $4.4670\times 10^{-5} mol$ of Hydrogen gas = :

$2 g/mol\times 4.4670\times 10^{-5} mol=8.934\times 10^{-5} g$

$9.7\times 10^{17}$ atoms of iron weigh as much as 1 cc of H2 at S.T.P.

Mass of $9.7\times 10^{17}$ atoms of iron =$8.934\times 10^{-5} g$

Moles of iron =$\frac{9.7\times 10^{17}}{6.022\times 10^{23} mol^{-1}}=1.6107\times 10^{-6} mol$

Atomic mass of iron: $\frac{\text{Mass of given atoms}}{Moles}$

:$\frac{8.934\times 10^{-5} g}{1.6107\times 10^{-6} mol}=55.49 g/mol$

The atomic mass of iron is 55.49 g/mol.

Answer 2

Answer:55.9 g

Explanation:mass of 1 cc of H2 at STP= 2.016÷22400 = 9 ×10^-5

Therefore 9.7 × 10^17 = 9×10^-5g

Also we know I mole = atomic mass = Avogadro number

Therefore atomic mass = 9×10^-5 ×6.02×10^23 ÷ 9.7×10^17 = 55.9g