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The expression relating molarity (M) and molality (m) or a solution is:

(A) $m = \large\frac{M}{\rho+M_1M_2} \quad$ (B) $m = \large\frac{M}{\rho-M_1M_2} \quad$ (C) $m = \large\frac{\rho+M_1M_2} {M} \quad$ (D) $m = \large\frac{\rho-M_1M_2} {M} \quad$
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Answer: (B) $m = \large\frac{M}{\rho-M_1M_2} \quad$
$m = \large\frac{n_2}{m_1}$
$m_1 = m_tot - m_2 = m_tot - \large\frac{m_2}{M_2} $$\times M_2$ $ = \rho V - n_2M_2$
Substituting, $m =\Large\frac{ \large\frac{n_2}{V} } {\rho - \large\frac{n_2}{V} \times M_2 }$$ =\large\frac{M}{\rho-M_1M_2} $
answered Jul 15, 2014 by balaji.thirumalai

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