Question

a) How many moles of Magnesium would be present in a Magnesium ribbon weighing 0.48 g ? (Atomic mass of Mg = 24 u)
OR
a) What is the no. of moles of Calcium that would be present in a sample of Calcium containing 1.2044 x 1020 atoms of Ca ? (Atomic mass of Ca = 40 u)

b) Calculate the formula unit mass of Sodium sulphite. (Atomic mass of Na = 23 u , S = 32 u , O = 16 u)
OR
b) Calculate the molar mass of Potassium nitrate.
(Atomic mass of K = 31 u , N = 14 u , O = 16u)

c) Name the compound represented by the formula (NH4)2SO4​

Answer 1

atomic mass of sodium= 23 g/mol

moles of sodium atoms in 46 gm= 46/23 = 2 moles

as mentioned in question, no of Na atoms = no of atoms of Ca atoms.

therefore no of atoms of ca atoms= 2 moles

therefore mass of 2 moles of Ca atoms = 2 x atomic wt of Ca = 2 mol x 40 g/mol = 80 g

Therefore answer is 80g.