Question

A solution has 28 gram oxalic acid in 275 ml. solution. And its density is 1.29g/ml
Calculate molality of the solution.
Or
46 gram of NaOH dissolved in 104gram of water if density of the solution is 1.48g/ml
calculate molarity of the solution​

Answer 1

Concepts used -

=> The concept of molality is used here where -

$\large{\boxed{Molality\:(in\:\:m)=\dfrac{No.\:of\:moles\:of\:solute}{No.\:of\:kilograms\:of\:solvent} }}$

=> The concept of Molarity is also used here where -

$\large{\boxed{Molarity\:(in\:\:M)=\dfrac{No.\:of\:moles\:of\:solute}{No.\:of\:litres\:of\:solvent} }}$

=> The concept of molar mass and number of moles is also used here, molar mass is equal to that mass of the compound which contains 1 mole of it.

Some things that we will need -

=> Here, I will find out and show some other values that we will need while finding the molarity and the molality -

• Molar mass of Oxalic acid -

   => Oxalic acid has the formula H2C2O4.2H2O

   => H = 1, C = 12, O = 16

   => 1*2 + 2 * 12 + 4 *16 + 4 * 1 + 2 * 16

  => 2 + 24 + 64 + 4 + 32

  => 30 + 64 + 32

  => 126 grams per mole

• Molar mass of NaOH -

       => NaOH (Sodium Hydroxide) -

       => Na = 23 , O = 16, H = 1

       => 23 + 16 + 1

       => 24 + 16

       => 40 grams per mole

#1 -

Evaluating -

∵ 28 grams Oxalic acid (H2C2O4.2H2O) is used

→ Molar mass of oxalic acid = 126 grams per mole.

→ Thus, 28 grams of H2C2O4.2H2O = $\dfrac{28}{126}$ => 0.222 moles

∵  0.222 moles of H2C2O4.2H2O is dissolved on 275 mL solution of            

  density 1.29 g/mL

∵ $\boxed{Density = \dfrac{Mass}{Volume}}$

→ 1.29 = $\dfrac{Mass}{275}$

→ Mass of solution = 1.29 * 275 g = 354.75 grams of solution

→ Out of this, 28 grams is oxalic acid -

→ So, mass of solvent (water) in grams = 354.75 - 28 => 326.75 grams

→ Mass of solvent in kilograms = $\dfrac{326.75}{1000}\:kg$ => 0.32675 kilograms

∵  $\boxed{Molality =\dfrac{No.\:of\:moles\:of\:H2C2O4.2H2O}{No.\:of\:litres\:of\:water}}$

→ Molality = $\dfrac{0.222}{0.32675}$

→ Molality = 0.679 m.

#2 -

Evaluating -

∵ 46 grams Sodium Hydroxide (NaOH) is used

→ Molar mass of NaOH = 40 grams per mole.

→ Thus, 46 grams of NaOH = $\dfrac{46}{40}$ => 1.15  moles

∵  46 grams of NaOH is dissolved on 104 grams of water.

∵ Density of solution = 1.48 g/mL

∵ $\boxed{Density = \dfrac{Mass}{Volume}}$

→ Mass of solution = 46 + 104 => 150 grams of solution

→ 1.48 = $\dfrac{150}{Volume\:in\:mL}$

→ Volume of solution = $\dfrac{150}{1.48}$ = 101.35 mL of solution

→ Volume of solution in Litres = $\dfrac{101.35}{1000}$ => 0.10135 L

∵  $\boxed{Molarity =\dfrac{No.\:of\:moles\:of\:NaOH}{No.\:of\:litres\:of\:solution}}$

→ Molarity = $\dfrac{1.15}{0.10135}$

→ Molarity = 11.346 M (molar)

Answers -

=> The molality of the Oxalic acid solution is 0.679 m .

=> The molarity of the NaOH solution is 11.346 M (molar)

More to know -

=> Learn more about Molarity, number of moles and co-relating it with chemical equations here -

https://brainly.in/question/37974785

Answer 2

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