Question

In an experiment it showed that 10 mL of 0.05 M solution of
chloride required 10 mL of 0.1 M solution of AgNO₃, which
of the following will be the formula of the chloride (X stands
for the symbol of the element other than chlorine):
[NEET Kar. 2013]
(a) X₂Cl (b) X₂Cl₂
(c) XCl₂ (d) XCl₄

Answer 1

(C)XCL2

• No. of moles of AgNO3=10-3 mol
• No. of moles the chloride= 0.5 x 10-3 mol
• Suppose the formula for the chloride is XCln then moles of chloride ion=n x 0.5 x 10-3
• Then, going by stoichiometry we get  n x 0.5 x 10-3= 10-3
• So n=2
• Therefore, formula is XCl2

Answer 2

The formula of the chloride is  XCl₂

Explanation:

From the given data, the chemical reaction involved in the experiment is $X C l_{x}+ A g N O_{3} \rightarrow A g C l+X nitrate$

10 mL, 0.05M $XCl_x$

10mL, 0.1M $AgNO_3$

According to molarity definition

$Molarity=\frac{No.\,of\,moles}{Volume}$

$No.\,of\,moles(n)=Molarity\times V$

$n_{AgNO_3}=0.1\times 10 \times 10^{-3} = 10^{-3}$$n_{chloride}=0.05\times10\times10^{-3}=5\times10^{-4}$

From the given data,

$5\times10^{-4} moles\, of \,chloride = 10^{-3} moles\,of\, AgNO_3$

Therefore, $1 mole\,of\,chloride=\frac{1 \times 10^{-3}}{5 \times10^{-4}} =2 moles\,of\, AgNO_3$

Therefore, the equation can be written as follows

$X C l_{2}+2 A g N O_{3} \rightarrow 2 A g C l+X\left(N O_{3}\right)_{2}$

Hence, the formula of the chloride is XCl₂

Therefore, the correct answer is option (c)