Question

Q-(I) Find the weight of 4*10(26) molecule of NH3 (26 in exponent and 3 in NH at base)
(II) If the weight of 1 atom of an element 'x'is 1.57*10(-24) gram. Find
the no. Of moles in 40kg of it? (-24is an exponent)

Answer 1

1)Weight of 1 mole of NH3 = 14+3=17g
1 mole NH3 contains 6.022 *10(23) molecules of NH3
So weight of 1 molecule NH3 = 17/6.022*10(23) =2.82 *10(-23) g
Weight of 4*10(26) molecules of NH3 = 2.82*10(-23) x4*10(26) =11.28*10(3) g

2)weight of 1 atom of an element X=1.57*10(-24)g
Number of atoms in 1g of elemrnt X=1/1.57*10(-24)
=0.64*10(24) atoms
So number of atoms in 40kg of element X =40*10(3) x 0.64*10(24) =25.6*10(27) atoms

(Numbers given in brackets () are exponents)