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An aqueous solution of $98 \%(w / w) \mathrm{H}_2 \mathrm{SO}_4$ has density of $1.02 \mathrm{~g} / \mathrm{cc}$. The molality of the solution is
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The correct answer is:
1.1
$\because$ Volume $=\frac{\text { Mass }(m)}{\text { Density }(d)}=\frac{100}{1.02}=98.039 \mathrm{~mL}$
$\therefore \quad$ Molarity $=\frac{9.8 \times 1000}{98 \times 98.04}=1.01$
Which is close to 1.1 .
$\therefore \quad$ Molarity $=\frac{9.8 \times 1000}{98 \times 98.04}=1.01$
Which is close to 1.1 .
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