Question

calculate molality of aq. solution of nitric acid obtained by dissolving 6.3 gram of HNO3 in 200 gram of H2O​

Answer 1

★$\color{darkblue}\underline{\underline{\sf Given-}}$

• Mass of nitric acid (m) = 6.3g
• Amount of solvent ${\sf (H_2O)}$ =200g

$\underline{\sf Molecular \: Weight-}$

$\implies{\sf HNO_3= 1+7+3(16) }$

$\color{orange}\implies{\sf HNO_3=63 }$

★$\color{darkblue}\underline{\underline{\sf To \: Find-}}$

• Molality of aqueous Solution of nitric acid.

★$\color{darkblue}\underline{\underline{\sf Formula \: Used-}}$

$\color{violet}\bullet\underline{\boxed{\sf Molality (m)=\dfrac{Mole \: of \: Solute}{Amount \: of \: solvent\:(g)}}}$

★$\color{darkblue}\underline{\underline{\sf Solution-}}$

Mole of ${\sf HNO_3}$ -

$\underline{\boxed{\sf Mole(n)=\dfrac{Mass}{Molecular \: mass}}}$

$\implies{\sf n=\dfrac{6.3}{63} }$

$\color{orange}\implies{\sf n=10 }$

$\large\implies{\sf Molality =\dfrac{Mole\: of \: Solute}{Amount \: of \: solvent}}$

$\implies{\sf Molality (m)=\dfrac{10}{200} }$

$\implies{\sf m=0.05m }$

★$\color{darkblue}\underline{\underline{\sf Answer-}}$

Molality is $\color{red}{\sf 0.05m}$