Question

in %m/m of a solution containing 39.2 g of potassium nitrate in 177 g of water ?​

Answer 1

★ Given :-

• Mass of solute (potassium nitrate - KNO3 )= 39.2 g
• Mass of solvent (water ) = 177 g

★ To find :-

• m /m %

★ Solution :-

As per the formula ,

$\bigstar\boxed{\mathtt{\% \dfrac{m}{ m}= \dfrac{m_{solute}}{m_{solute}+ m_{solvent }} \times 100}}$

Now let's substitute the given data in the above equation ,

$\implies\mathtt{\% \dfrac{m}{ m}= \dfrac{39.2}{39.2+177} \times 100}$

$\implies\mathtt{\% \dfrac{m}{ m}= \dfrac{39.2}{216.2} \times 100}$

$\implies\mathtt{\% \dfrac{m}{ m}= 0.18 \times 100}$

$\implies \boxed{\mathtt{\% \dfrac{m}{ m}= 18\%}}$

★Additional Information:-

$\implies\:\mathtt{\% \dfrac{v}{ v}= \dfrac{v_{solute}}{v_{solute}+ v_{solvent }} \times 100}$

$\implies\:{\mathtt{\% \dfrac{m}{ v}= \dfrac{m_{solute}}{v_{solute}+ v_{solvent }} \times 100}$

$\implies\:\mathtt{Molarity = \dfrac{moles \:of \:solute }{volume\: of\: solution} }$

$\implies\:\mathtt{Molality = \dfrac{moles \:of \:solute }{mass\: of\: solvent(kg)} }$