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Q.

Density of 2.05M solution of acetic acid in water is 1.02 g/mL, the molality of that solution is

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a

0.43

b

0.06

c

0.49

d

2.3

answer is A.

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Detailed Solution

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$\large molality=\frac{MolarityX1000}{(1000Xd\;of\;sol.)-(MolarityXMW\;of\;solute)}$

$\large =\frac{2.05X1000}{(1000X1.02)-(2.05X60)}=\frac{2050}{1020-123}=\frac{2050}{897}$

$\large =2.3m$

Alternate method:

$\begin{array}{l} {M_{C{H_3}COOH}} = 2.05M,\,\,\,\,{n_{C{H_3}COOH}} = 2.05\\ {V_{sol.}} = 1000ml,\,\,\,\,\,\,\,{d_{sol.}} = 1.02g/cc\\ {W_{sol.}} = 1000(1.02) = 1020g \end{array}$

$\begin{array}{l} {W_{C{H_3}COOH}} = 2.05(60) = 123g\\ {W_{solvent}} = (1020 - 123) = 897g \end{array}$

$\begin{array}{l} molality = {n_{C{H_3}COOH}}X\frac{{1000}}{{{W_{solvent}}(g)}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2.05X\frac{{1000}}{{897}} = 2.3m \end{array}$

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Density of 2.05M solution of acetic acid in water is 1.02 g/mL, the molality of that solution is