$(a)\;3g/dm^3\qquad(b)\;4g/dm^3\qquad(c)\;5g/dm^3\qquad(d)\;6g/dm^3$

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Given

d(g) = 5.46 $3g/dm^3$

T = 300K

P = 2 bar

d(g) = ?

T = 273K

P = 1 bar

Since $\large\frac{d_1}{d_2} = \large\frac{P_1m_1}{RT_1}\times\large\frac{RT_2}{P_2m_2}$

($m_1 = m_2$ for same gas)

$\large\frac{5.46}{d_2} = \large\frac{2\times273}{1\times300}$

$1\times300\times5.46 = 2\times273\times d_2$

$\large\frac{300\times5.46}{2\times273} = d_2$

$d_2 = 3g/dm^3$

Hence answer is (a)

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