Question

The weight of one atom of uranium is 238 amu...what is its actual weight in grm

Answer 1

actual weight in grams is 397.46×10^_27

Answer 2

$\text{The actual weight in grams is} \ 3.97 \times 10^{-22} \ \mathrm{g}$

Given:

The weight of one atom of Uranium is 238 amu.

Solution:

$1 \ amu = 1.67 \times 10^{-27} \ \mathrm{kg}$

Therefore,

$238 \ \text { amu }=238 \times 1.67 \times 10^{-27} \ \mathrm{kg}=397.46 \times 10^{-27} \ \mathrm{kg}$

$238 \ \mathrm{amu}=3.97 \times 10^{-25} \ \mathrm{kg} \text { or } 3.97 \times 10^{-22} \ \mathrm{g}$

Alternative method:

Weight of one atom of Uranium = 238 amu

Weight of one mole of uranium atoms = 238 g  = 0.238 kg

Number of atoms in one mole of uranium $=6.023 \times 10^{23}$

Therefore,

$\text{Weight of one atom of uranium} =\frac{0.238}{6.023 \times 10^{-23}}$

$\text{Weight of one atom of uranium} =3.95 \times 10^{-25} \ \mathrm{kg}$