Question

A metal forms two oxides. The higher oxide contains $80\%$ metal. $0.72g$ of the lower oxide gave $0.8g$ of higher oxide when oxidized. Then the ratio of the weight of oxygen that combines with the fixed weight of the metal in the two oxides will be:

• A. $2:3$
• B. $2:1$
• C. $4:5$
• D. $3:2$

Answer 1

The correct option is B $2:1$

Given that ,in higher oxide 80% metal is true.

wt. of metal$=\frac{80}{100}\times 0.8$

$=0.64gm$

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 = $0.75\times 25/1000$ = 0.01875

2 mol of HCl reacts with 1 mol of $CaC{O}_3$

So, 0.01875 mol of HCl will react with $\frac{1}{2}$ $\times$ 0.01875 = $0.009375$ mol

Molar mass of $CaC{O}_3$ = 100 g

Hence, the mass of 0.009375 mol of $CaC{O}_3$ = no. of moles $\times$ molar mass$

$0.009375$ $\times 100$

Hence, the correct option is $C$.