Question

Two oxides of an metal contains 80% and 88.8% of the metal respectively. show that it is multiple Proportion .
​

Answer 1

Given that ,in higher oxide 80% metal is true.
wt. of metal=
100
80
​
×0.8

=0.64gm
wt. of oxygen=0.16gm
in lower oxide,
wt. of metal will be same as that of higher oxidation state(+2)

wt. of oxygen=(72−0.64)=0.08g

so,ratio of weight of O
2
​
& meals in two oxide is 0.16:0.08=2:1.
1000ml of 0.75 M HCl = 0.75 mol

So, 25 ml of HCl will contain HCl = 0.75×25/1000 = 0.01875

2 mol of HCl reacts with 1 mol of CaCO
3
​


So, 0.01875 mol of HCl will react with
2
1
​
× 0.01875 = 0.009375 mol

Molar mass of CaCO
3
​
= 100 g

Hence, the mass of 0.009375 mol of CaCO
3
​
= no. of moles × molar mass$$