Question

the number of atoms present in 1.0cm^3 of solid glucose (density 0.8/cm^3)

Answer 1

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See,Density=Mass/Volume



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Answer 2

The number of atoms present in 1.0cm^3 of solid glucose is 27× 10^20 atoms.

Given:

1.0cm^3 of solid glucose of density 0.8g/cm^3.

To Find:

The number of atoms present in 1.0cm^3 of solid glucose.

Solution:

To find the number of atoms present in 1.0cm^3 of solid glucose we will follow the following steps:

As we know,

$density = \frac{mass}{volume}$

$mass = volume \times density = 1.0 \times 0.8 = 0.8gram$

$moles \: of \: glucose = \frac{given \: mass}{molar \: mass} = \frac{0.8}{180} = 0.0044$

The molecular weight of glucose = 180gram.

Several atoms present in glucose = moles × Avogadro number of particles = 0.0044× 6.022×10^23 = 0.027× 10^23 = 27× 10^20 atoms.

Henceforth, the number of atoms present in 1.0cm^3 of solid glucose is 27× 10^20 atoms.

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