$(a)\;CH_3Cl\qquad(b)\;C_2H_4Cl_2\qquad(c)\;C_3H_{12}Cl_5\qquad(d)\;CH_3Cl_2$

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Step 1:

$PV=\large\frac{w}{m}$$RT$

$\large\frac{768}{760}$$\times \large\frac{37.24}{1000}=\large\frac{0.120}{m}$$\times 0.0821\times 378$

$m=99$

Let formula of compound be $C_aH_6Cl_2$

$C_aH_bCl_2+(a+\large\frac{b}{4})$$O_2\rightarrow aCO_2+\large\frac{b}{2}$$H_2O+\large\frac{z}{2}$$Cl_2$

$99g\Rightarrow 44ag\;CO_2$

$6.22g\Rightarrow \large\frac{44a\times 0.22}{99}$

$\therefore \large\frac{44a\times 0.22}{99}$$=0.195$

$a=2$

Step 2:

Similarly $0.22g$ compound gives =$\large\frac{18\times b\times 0.22}{2\times 99}$g $H_2O$

$\large\frac{18\times b\times 0.22}{2\times 99}$$=0.0804$

$b=4$

Now $C_aH_bCl_2$

$12\times a+b+35.5z=99$

$24+4+3.5z=99$

$z=2$

$\therefore C_2H_4Cl_2$

Hence (b) is the correct answer.

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