Question

calculate molality of 1 M solution of NaNO₃ if its density is 1.25g/cm³

Answer 1

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Answer 2

$\textsf{\underline {\Large {Molality}}}$ :

Given, Moles of NaNO₃ = 1

Density = 1.25 g/cm³

$\boxed{\mathsf{Molality \:=\:{\dfrac{Number \:of\:moles\:of\:solute\:}{Mass\:of\:the\:solvent(\:kg\:) }}}}$

Molar Mass of NaNO₃ = 23 + 14 + 16× 3 = 37+ 48 = 85 g

Mass of NaNO₃ in 1 L solution = 1 × 85 = 85 g

Density = $\mathsf {\dfrac{\:Mass} {Volume}}$

1.25 = $\mathsf {\dfrac{\:m} {1000}}$

↪️Mass of 1 L solution = $\mathsf {1000\:{\times{1.25}}\:=\:1250\:g}$

Mass of water in solution = 1250 - 85 = 1165 g

Putting value of " volume" and " moles" in formula of Molarity,

$\mathsf{Molality \:=\:{\dfrac{Number \:of\:moles\:of\:solute\:}{Mass\:of\:the\:solution(\:kg\:) }}}$

$\mathsf{Molality \:=\:{\dfrac{1}{1165}}}$

$\mathsf{Molality \:=\:0.000858}$

$\boxed{\mathsf{Molality \:=\:8.58\:{\times{{10}^{-4}}m}}}$