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Question: An atom of an element weigh $6.4 \times 10^{-23}$ g. Number of gram atoms present in 30 Kg of it?...

An atom of an element weigh 6.4×10−236.4 \times 10^{-23} g. Number of gram atoms present in 30 Kg of it?

A

650

B

700

C

778

D

540

Answer

778

Explanation

Solution

Here's how to solve this problem:

  1. Calculate the atomic mass (molar mass) of the element: Multiply the mass of one atom (6.4×10−236.4 \times 10^{-23} g) by Avogadro's number (NA≈6.022×1023N_A \approx 6.022 \times 10^{23} mol−1^{-1}). This gives the mass of one mole of atoms, which is approximately 38.54 g/mol.

    Atomic mass=(6.4×10−23g)×(6.022×1023atoms/mol)≈38.54g/molAtomic\ mass = (6.4 \times 10^{-23} g) \times (6.022 \times 10^{23} atoms/mol) \approx 38.54 g/mol

  2. Convert the total given mass from kilograms to grams: 30 Kg = 30000 g.

  3. Calculate the number of gram atoms (moles): Divide the total mass (30000 g) by the atomic mass (38.54 g/mol).

    Number of gram atoms=30000 g38.54 g/mol≈778.38 molNumber\ of\ gram\ atoms = \frac{30000\ g}{38.54\ g/mol} \approx 778.38\ mol

  4. Round to the nearest integer: 778.38 ≈ 778