Question

Calculate the number of molecules present in 34.20 grams of cane sugar

Answer 1

Solution :

1 mole of C12H22O11=342gC12H22O11=342g

= 12 x 12 + 22 x 1 + 11 x 16 = 342 amu

=6.022×1023=6.022×1023 molecules

Now 342 g of cane sugar contain 6.022×10236.022×1023 molecules.

34.2 g of cane sugar will contain

=6.022×1022=6.022×1022 molecules

Answer 2

Given - Mass of cane sugar

Find - Number of molecules

Solution - 1 mole of any compound contains Avogadro's number of molecules.

Number of moles = mass/molar mass

Number of moles = 34.2/342

Number of moles = 0.1

Number of molecules in 1 mole of cane sugar = 6.02*10²³

Number of molecules 0.1 mole of cane sugar = 0.1/6.02*10²³

Number of molecules = 6.02*10²⁴

Hence, the number of molecules present in 34.20 grams of cane sugar is 6.02*10²⁴.