$(a)\;12\qquad(b)\;11.22\qquad(c)\;10.3\qquad(d)\;12.5$

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Step 1:

Let $V$ml of each are mixed.

$H_2SO_4$ is 30% by weight

Wt of $H_2SO_4=30g$

Wt of solution =100g

Volume of solution=$\large\frac{100}{1.218}$$ml$

(i.e) $\large\frac{100}{1.218}$$ml$ contains 30g $H_2SO_4$

$\therefore$ Vml contains $\large\frac{30\times V\times 1.218}{100}$ of $H_2SO_4$

Step 2:

$H_2SO_4$ is 70% by wt.

$\therefore$ Wt of $H_2SO_4$=70g

Wt of solution =100g

Volume of solution =$\large\frac{100}{1.610}$ml

(i.e) $\large\frac{100}{1.610}$ml contains 70g $H_2SO_4$

$V$ml contains $\large\frac{70\times V\times 1.610}{100}$g $H_2SO_4$

On mixing these two,total wt of $H_2SO_4=\bigg[\large\frac{30\times 1.218}{100}+\frac{70\times 1.610}{100}\bigg]$$Vg$

$\Rightarrow 1.4924Vg$

Step 3:

Total volume of solution =2Vml

Molarity of solution =$\large\frac{1.4924V}{98\times \Large\frac{2V}{1000}}$

$\Rightarrow 7.61$

Now wt of total solution $=2V\times 1.425g$

$\Rightarrow 2.85Vg$

Wt of water =(2.85V-1.4924V)g

$\Rightarrow 1.3576Vg$

Molality of solution =$\large\frac{1.4924V}{98\times \Large\frac{1.3576V}{1000}}$

$\Rightarrow 11.22$

Hence (b) is the correct option.

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