Question

Calculate the number of chromium atom in 1.47 g of pota ium dichromate (K2Cr2O7) [Atomic mass: K=39, Cr=52, O=16].​

Answer 1

Answer:

The number of chromium atoms in 1.47 grams of potassium dichromate

                                                                               = 6.022×10²¹ atoms.

Explanation:

The chemical formula of potassium dichromate = K₂Cr₂O₇

Potassium dichromate has 2 atoms of chromate atoms in each molecule.

The atomic mass of potassium = 39$g/mol$

The atomic mass of chromium= 52$g/mol$

The atomic mass of oxygen = 16$g/mol$

Hence, the molecular mass of potassium dichromate = (39×2)+(52×2)+(16×7)

                                                                                   = 78+104+112

                                                                                    = 294$g/mol$

one mole of potassium dichromate has 2 moles of chromium atoms in it,

so, 294-grams K₂Cr₂O₇ has 104 grams of chromium atom.

∴ amount of chromium atom in one gram K₂Cr₂O₇ = $\frac{104}{294}$ = 0.35 gram

Therefore the amount of chromium in 1.47 grams of K₂Cr₂O₇  = 1.47×0.354

                                                                                         = 0.52 grams

Number of moles of chromium atoms here = $\frac{0.52}{52}$ = 0.01 moles

∴ number of chromium atoms in 1.47 grams  of potassium dichromate=

                                                                      = 0.01×6.022×10²³

                                                                       = 6.022×10²¹ atoms.

Answer 2

Calculate the number of chromium atom in 1.47 g of pota ium dichromate (K2Cr2O7) [Atomic mass: K=39, Cr=52, O=16].

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Molecular mass of K2 Cr 2 O7

= 2×39 + 2×52 + 16×7

= 78+104+112

= 294

There are number of Cr atom in 294 gm .

K2 Cr2 O7 = 2×6.022×10²³

No of Cr atom in 1.47 gm K2 Cr2 O7

= 2×6.022×10²³×1.47 / 294

= 6.022×10²¹ atoms

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