A compound on analysis is found to contain Na = 17.03%, S = 47.40% and O = 35.55%. If the molecular mass of the compound is 270, then the number of sulphur atoms present in one molecule of the compound is :

Ratio of percentage of Na, S and O is 17.03 : 47 .40 : 35.55∴Ratio of atoms of Na, S and O is=17.0323:47.4032:35.5516=0.74:1.48:2.22=1:2:3E.F. of the compound: NaS2O3 E.F. Mass =23+2×32+16×3=135∴ M.F. =(NaS2O3)×270135=Na2S4O6Thus, the no. of atoms of S per molecule = 4.