Given,

weight of sodium =0.46 gm

weight of sodium salt =1.64 gm

equivalent weight of sodium =23

equivalent weight of acid =?

We Know,

`E_(Na)/E_(Na.sal t)=W_(Na)/W_(Na.sal t)`

or,`23/(23+x)=0.46/1.64`

or,`x=59 gm`

Now, equivalent weight of an acid,HX `=59+1=60`

We Know,

1 mole of `H_2O=N_A` `H_2O` atoms=18 gm
Mass of 1 ml of `H_2O=1**1=1gm`
1 gm of `H_2O=N_A/18 H_2O` atoms
`=N_A/18**2` H-atoms
`=6.69**10^22` H-atoms

We Know,

1 mole of `H_2O=18g m=6.023**10^23` molecules

Mass of water in 1 ml=`rho**V=1**1=1 `g/cc

Now,

18 gm of water has `6.023**10^23` `H_2O` molecules.

1 gm water has `(6.023**10^23)/18=3.34**10^23` `H_2O` Molecules

 We Know,

1 mole of `NH_3`=22.4 l of `NH_3=N_A` `NH_3` molecules

Or, 5.6 l of `NH_3=N_A/22.4**5.6\ NH_3` molecules
`=4**N_A/22.4**5.6` individual atoms
`=N_A` atoms

Molecular weight of metal `=20**3=60`

`4M+3O_2->2M_2O_3`

Thus, Molecular weight of metal oxide `=60**2+16**3=168`

Given,

weight of acid =1 gm

weight of magnesium salt =1.301 gm

equivalent weight of magnesium =12

equivalent weight of acid =?

We Know,

`E_(acid)/E_(mg.sal t)=W_(acid)/W_(mg.sal t)`

or,`(1+x)/(12+x)=1/1.301`

or,`x=35.5 gm`

Now, equivalent weight of an acid,HX `=35.5+1=36.5`

We know,

Number of mole of HCl =Molarity *Volume of solution in liters

`=(1**200)/1000=0.2\ mol=7.3\ gm`

The required reaction is,

`CaCO_3+2HCl->CaCl_2+H_2O+CO_2`

From above balanced equation, We get,

100 gm of `CaCO_3` reacts with 73 gm of `HCl` to give 111 gm of `CaCl_2`

or, 5 gm of `CaCO_3` reacts with 3.65 gm of `HCl` to give 5.55 gm of `CaCl_2` and excess that remains is `HCl`.

Eq. wt of Metal`=(Wt.of\ met al)/(Wt.of\ chlo ri n e)**35.5`

`=(100-37.322)/37.322**35.5`

`=59.61`

Given,

`1l` of metal chloride vapour at STP `=16.96gm`

We Know,

`22.4l` of metal chloride at STP = molecular weight of metal chloride

or,`22.4l=`valency `**` equivalent weight of metal chloride

or, `22.4l=Val ency**(59.61+35.5)`

or,`1l=Val ency**95.11/22.4=16.96(given)`

or, `val ency=3.99~=4`

Thus, formula of metal chloride is `MCl_4`

We Know,

`(eqv.wt.of\ met al\ chlo ride)/(eq.wt.of\ me tal\ o x ide)=(wt.of\ met al\ chlo ride)/(wt.of\ me tal\ o x ide)`

let, equivalent weight of metal=x.

or,`(x+35.5)/(x+8)=5/3`

or,`x=(3**35.5-5**8)/(5-3)`

or,`x=33.25`