Question

Two oxide of metal contain 80% and 88.8% of the metal respectively.Show that the data in agreement with multiple proportion.​

Answer 1

Answer:

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$$

0.009375 ×100