Question

Molarity of 29%(w/w) H2SO4 solution whose density is 1.22 g/ml is

Answer 1

Answer:

3.61M

Explanation:

mol wt of h2so4 =98

now 29% means

100g solution contains 29 gm h2so4

i,e 100/1.22 ml contains 29/98 moles(as d=1.22g/ml)

1000ml contains 29*1.22*10/98=3.61 M

Answer 2

Final answer: Molarity of  $29$ % ($\frac{w}{w}$) $H_{2}SO_{4}$ solution whose density is is $1.22$ $\frac{g}{mL}$ is $3.6085$ $M$.

Given that: We are given $29$ % ($\frac{w}{w}$) $H_{2}SO_{4}$ solution whose density is $1.22$ $\frac{g}{mL}$.

To find: We have to find molarity of $H_{2}SO_{4}$ solution.

Explanation:

• We are given that $\frac{w}{w}$ % of sulphuric acid.

                $\frac{w}{w}$ % of sulphuric acid = $29$ %

Which means that $29$ $g$ of given sulphuric acid is dissolved in $100$ $g$ of solution.

• The equation for molarity ($M$),

           $Molarity, (M)=\frac{Number of moles of solute}{Volume of solution in litre}$

           $Molarity of H_{2}SO_{4} solution (M)=\frac{Number of moles of H_{2}SO_{4}}{Volume of H_{2}SO_{4} solution in litre}$

• $Number of moles = \frac{Given mass (g)}{Molar mass (\frac{g}{mol})}$

         $Number of moles of H_{2}SO_{4}= \frac{Given mass of H_{2}SO_{4} (g)}{Molar mass of H_{2}SO_{4} (\frac{g}{mol})}$

         Given mass of $H_{2}SO_{4}$ $= 29$ $g$

         Molar mass of $H_{2}SO_{4}$ $= 98$ $\frac{g}{mol}$

         Number of moles of $H_{2}SO_{4}$ = $\frac{29}{98}$ $=0.2959$ $mol$

• $Volume of H_{2}SO_{4} solution = \frac{Mass of H_{2}SO_{4} solution}{Density of H_{2}SO_{4} solution}$

• Density of given sulphuric acid solution =  $1.22$ $\frac{g}{mL}$

          Mass of $H_{2}SO_{4}$ solution $= 100$ $g$

          Volume of $H_{2}SO_{4}$ solution = $\frac{100}{1.22}$ $=81.967$ $mL$

          Volume of $H_{2}SO_{4}$ solution in litre = $\frac{81.97}{1000}$ = $0.08197$ = $0.082$ $L$

• Substitute, number of moles of $H_{2}SO_{4}$ and volume of $H_{2}SO_{4}$ solution in litre in molarity of  $H_{2}SO_{4}$ solution.

          Molarity of $H_{2}SO_{4}$ solution ($M$)= $\frac{0.2959 mol}{0.082 L}$=$3.6085$ $M$

• Hence, molarity of  $29$ % ($\frac{w}{w}$) $H_{2}SO_{4}$ solution whose density is is $1.22$ $\frac{g}{mL}$ is $3.6085$ $M$.

To know more about the concept please go through the links

https://brainly.in/question/5266762

https://brainly.in/question/3626950

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