2.atomic weight of y is options.70;20;40;65
3.no.of x atoms present in compound is 
a] 5.01 x 10 power22      b] 6.02 x 10 power21    c] 3.01 x 10 power 23    d] 1.204 x 10 power 22

Dear Student,

Please find below the solution to the asked query 1:

the empirical formula of the compound is : xy₂z₃ 

Now, as we know that not all the x could have been used in making the compound, as we are given 5 g of x and the total mass of product is = 4.40 g. Thus it must be present in an excess amount. 

Moles of y = $\frac{1.15 x {10}^{23} atoms}{6.022 x {10}^{23} atoms mo{l}^{-1}} = 0.19 mol$


As we are told that there are 0.03 moles of z, then it must be a limiting reagent,
hence, we can assume that all of one of the elements is in the compound which has to be z. 

Thus, assuming there are 0.03 moles of z in the compound, then using the given empirical formula:


0.02 moles y and 0.01 moles x

Molar mass of x = 60 g mol^-1 
Thus mass if x = 60 g mol^-1  x 0.01 = 0.6 g 

Similarly, molar mass of z = 80 g mol^-1 
Thus mass of z = 80 g mol^-1 x 0.03 mol = 2.4 g 

thus mass of y = total mass of the compound - (mass of x + mass of y) 
                      y =   total mass of the compound - mass of x - mass of y
 
                            = 4.40 g - 0.6 g - 2.4 g = 1.4 g 


Since the moles of y = 0.02 mol

atomic mass of y = $\frac{1.4 g}{0.02 mol}$ = 70 g mol^-1 
Hence the molecular weight of the compound =  xy₂z₃ =   60 + (2 x 70) + (3 x 80) g mol^-1
                                                                               =  (60  +  140  +  240) g mol^{-1
                                                                                               =}   440 g mol^-1
Hence option (B) is correct


For the remaining questions please ask in diffrent thread