Identify the incorrect statement(s)

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Molarity of very dilute solution with CC1₄ as solvent is greater than its molality if density of $C C l_{4}$ is $1.6 g / c m^{3} .$

An oleum with labelling $\frac{196}{178} \times 100$ will have mass % combined SO₃ equal to $\frac{160}{178} \times 100 % .$

The value of universal gas constant is (0.0821 × 0.76) in mm of $H g - m^{3} / m o l e$ Kelvin.

For all gaseous reaction if volume changes during reaction at constant temperature and pressure then vapor density may remain constant.

answer is B, D.

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Detailed Solution

(A) Molarity = $\frac{M o l e}{L i t e r}$
In case of very dilute solution Volume of solution = Volume of solvent
molality = mole /kg
1 kg of solvent = 1000 gram as density of solvent $C C l_{4}$ = 1.6 g/mL
Volume of solvent = $\frac{1000}{1.6} m l$
molarity > molality
(B) $\frac{196}{178} \times 100 %$ OLEUM
178 gm oleum means 80g $S O_{3}$ and 98 $H_{2} S O_{4}$
% of combine $S O_{3} = \frac{\frac{80}{98} \times 100}{178}$
(C)    $0.0821 L a t m K^{- 1} m o l^{- 1}$
   $0.0821 \times 10^{- 3} m^{3} \times 760 m m H g K^{- 1} m o l^{- 1} \begin{array}{l} 0.0821 \times 10^{- 3} \times 760 H g m^{3} m o l^{- 1} K^{- 1} \\ 0.0821 \times 0.76 H g m^{3} m o l^{- 1} K^{- 1} \end{array}$

(D) Volume change as $Δ n g \neq O$ So vapour density will change A,C are correct and BD are incorrect