(C) The relationship between molarity $(M)$,molality $(m)$,and density $(d)$ is given by the formula: $m = \frac{1000 \times M}{1000 \times d - M \tim

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(C) The relationship between molarity $(M)$,molality $(m)$,and density $(d)$ is given by the formula: $m = \frac{1000 \times M}{1000 \times d - M \times M_B}$ Rearranging for density $(d)$: $d = \frac{M}{m} + \frac{M \times M_B}{1000} = \frac{1.44}{1.5} + \frac{1.44 \times 166}{1000}$ $d = 0.96 + 0.23904 = 1.19904 \ g \ mL^{-1} \approx 1.20 \ g \ mL^{-1}$ Here,$M = 1.44 \ mol \ L^{-1}$,$m = 1.5 \ mol \ kg^{-1}$,and $M_B (KI) = 166 \ g \ mol^{-1}$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

Medium

Calculate the density of an aqueous solution of $KI$,if its molarity and molality are $1.44 \ M$ and $1.5 \ mol \ kg^{-1}$ respectively. (in $g \ mL^{-1}$)

AP EAMCET 2020 All AP EAMCET

A
$2.20$
B
$2.50$
C
$1.20$
D
$0.50$

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Calculate the density of an aqueous solution of $KI$,if its molarity and molality are $1.44 \ M$ and $1.5 \ mol \ kg^{-1}$ respectively. (in $g \ mL^{-1}$)

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