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Calculate the molarity and molality of a 15% solutions (by weight) of sulphuric acid density $1.020g cm^{-3}$.(Atomic masses H=10=16S=32 amu)

$\begin{array}{1 1}\text{1.3048 molar,1.53 molal}\\\text{2.3048 molar,2.53 molal}\\\text{3.3048 molar,3.53 molal}\\\text{4.3048 molar,4.53 molal}\end{array} $

1 Answer

$M=\large\frac{W_B}{M_B\times V_L}$
Molecular weight of $H_2SO_4=2\times 1+32+64$
$\Rightarrow 98g mol^{-1}$
$M=\large\frac{15g\times 1000}{98g/mol\times 115g \times 1.02g/cm^3}$
$\;\;\;\;=\large\frac{15000}{11495.4}$ molar
$\;\;\;\;=1.3048$ molar
m (molality) =$\large\frac{W_B\times 1000}{W_B \times 100g}$
$\Rightarrow \large\frac{15g\times 10}{98g mol^{-1}}$
$\Rightarrow \large\frac{150}{98}$
$\Rightarrow 1.53$ molal
answered Aug 8, 2014 by sreemathi.v
 

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