Total no. of atoms present in a box containing 7 gm atoms of C12H22O11 and 3 moles of NH3 would be.......................

 let ,no.of moles of c12h22o11 be A= 7/molecular mass of c12 h22o11

no.of C atoms= A* NA*12

no.of H atoms=A*NA*22(due to c12h22O11)

no.of O atoms=A*NA*11

no.of H atoms =3*NA*3(due to NH3)

no.of N atoms=9*NA*1

total atoms =sum of all the atoms

if this information really helps you, then plz reply yes or no.