$(a)\;1.64\qquad(b)\;1.88\qquad(c)\;1.22\qquad(d)\;1.45$

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Given that,mass $\%$ of $H_2SO_4=29\%$

(i.e) 100g solution contains 29g $H_2SO_4$

Let the density of solution (in g/ml) is d

$\therefore$ Molarity of solution =$\large\frac{Moles\;of\;H_2SO_4}{Volume\;of\;solution(in\;ml)}$$\times 1000$

$\Rightarrow \large\frac{29/98}{100/d}$$\times 1000=3.60$

(M=3.60)

$d=1.22gml^{-1}$

Hence (c) is the correct option.

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