Question

Calculate the total number of oxygen atoms present in 1.80g of glucose

Answer 1

molar mass of C6H12O6 =180g/mol
then, in 1.80g glucose number of O atom present = 1.80/180×6.022*10^23×6
=0.06×6.022*10^23
=3.6132*10^22

Answer 2

Answer:

no. of O atoms in glucose=

3.6132x10^-22atoms

Explanation:

concept- total no. of O atoms = no. of moles of oxygen*avogadro no.*no. of atoms of O in glucose

no. of mole = weight of solute/ molar mass

given - weight of solute= 1.80g

molar mass of glucose -180g

find = total number of oxygen atoms present

explanation-

number of moles- 1.80/180= .01moles

total no. of O atoms - no. of moles of oxygen*avogadro no.*no. of atoms of O in glucose

no. of O atoms= .01*6.022x10^-23*6

3.6132x10^-22atoms

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