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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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A $1.37\; M$ aqueous solution of citric acid ($H_3C_6 H_5O_7$; Molecular Weight = $192.1 g/mol$) has a density of $1.10 g/cm^{3}.$ What is the molality and mole fraction of the citric acid?

$\begin{array}{1 1} (A) molarity = 1.24 mol/kg, mole fraction = 0.028 \\ (B) molarity = 1.64 mol/kg, mole fraction = 0.029 \\ (C) molarity = 1.34 mol/kg, mole fraction = 0.039 \\ (D) molarity = 1.54 mol/kg, mole fraction = 0.049\end{array}$

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Let's assume that we have 1 L = 1000 mL of the solution.
1.37 mol $\times$ 192.1 g/mol = 263.18 g of citric acid.
Therefore, in 1 L, we will have 1000 ml $\times$ 1.1 g/cm$^3$ = 1.1 $\times$ 10$^3$ g solution.
$\Rightarrow$ 1.1 $\times$ 10$^3$ g solution - 263.18 g citric acid = 836.82 g $H_2O$
Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution.
$\Rightarrow$ molarity = 1.37 mol / 0.83682 Kg of $H_2O$ = 1.64 mol/kg.
836.82 g of $H_2O$ / 18.0 g/mol = 46.49 mol $H_2O$
$\Rightarrow$ mole fraction of citric acid = 1.37 mol / (1.37 mol + 46.49 mol) = 0.029.
answered Feb 13, 2014 by sreemathi.v
edited Mar 25, 2014 by balaji.thirumalai
 

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