250 cm3 of sulphur IC acid contain 24.5 g of h2so4. The density of sol

1.	250 cm3 of sulphur IC acid contain 24.5 g of h2so4. The density of solution is 1.98g/cm3. Determine molality
Answer» GIVEN : 250 cm³ sulphuric acid ( $\sf H_2 SO_4$ ) contain 24.5 g of $\sf H_2 SO_4$ . And, The DENSITY of the SOLUTION is 1.98 g/cm³. ⠀━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀ Formula of Molality (m) is given by, ⠀⠀ $\bigstar\:{\boxed{\sf{\pmb{Molality = \dfrac{No.\:of\:moles\:of\:Solute}{Mass\:of\:solvent\:in\:kg}}}}}\\\\\\$ So, First of all we will calculate Mass of Solvent – ⠀⠀ We're provided in the QUESTION that, the density of the solution is 1.98 g/cm³. So, From this we can easily calculate Mass of solution then Mass of Solvent. ⠀⠀ $\dashrightarrow\sf Density = \dfrac{Mass}{Volume}\\\\\\ \dashrightarrow\sf 1.98\:g/cm^3 = \dfrac{Mass_{\:(solution)}}{250\:cm^3}\\\\\\ \dashrightarrow\sf Mass_{\:(solution)} = 1.98 \times 250\\\\\\ \dashrightarrow{\boxed{\frak{Mass_{\:(solution)} = 495\:g}}}\\\\\\$ If Mass of Solution is 495 g and Mass of Solute is 24.5 g then, Mass of Solvent = 495 - 24.5 = 470.5 g = 0.4705 kg ⠀━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀ Hence, We've calculated Mass of Solvent i.e. 0.4705 kg. Now, Let's FIND out Molality — ⠀⠀ $\longrightarrow\sf m = \dfrac{No.\:of\:moles\:of\:Solute}{Mass\:of\:solvent\:in\:kg}\\\\\\ \longrightarrow\sf m = \dfrac{24.5}{98 \times 0.4705}\\\\\\ \longrightarrow\sf m = \dfrac{0.25}{0.4705}\\\\\\ \longrightarrow{\boxed{\sf{\pmb{\pink{m = 0.531\:molal}}}}}\:\bigstar\\\\\\$ $\therefore$ Hence, The Molality of Given solution is 0.531 m.

250 cm3 of sulphur IC acid contain 24.5 g of h2so4. The density of solution is 1.98g/cm3. Determine molality

Answer»

GIVEN : 250 cm³ sulphuric acid ( $\sf H_2 SO_4$ ) contain 24.5 g of $\sf H_2 SO_4$ . And, The DENSITY of the SOLUTION is 1.98 g/cm³.

⠀━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀

Formula of Molality (m) is given by,

⠀⠀

$\bigstar\:{\boxed{\sf{\pmb{Molality = \dfrac{No.\:of\:moles\:of\:Solute}{Mass\:of\:solvent\:in\:kg}}}}}\\\\\\$

So, First of all we will calculate Mass of Solvent –

⠀⠀

We're provided in the QUESTION that, the density of the solution is 1.98 g/cm³. So, From this we can easily calculate Mass of solution then Mass of Solvent.

⠀⠀

$\dashrightarrow\sf Density = \dfrac{Mass}{Volume}\\\\\\ \dashrightarrow\sf 1.98\:g/cm^3 = \dfrac{Mass_{\:(solution)}}{250\:cm^3}\\\\\\ \dashrightarrow\sf Mass_{\:(solution)} = 1.98 \times 250\\\\\\ \dashrightarrow{\boxed{\frak{Mass_{\:(solution)} = 495\:g}}}\\\\\\$

If Mass of Solution is 495 g and Mass of Solute is 24.5 g then,

Mass of Solvent = 495 - 24.5 = 470.5 g = 0.4705 kg

⠀━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀

Hence, We've calculated Mass of Solvent i.e. 0.4705 kg. Now, Let's FIND out Molality —

⠀⠀

$\longrightarrow\sf m = \dfrac{No.\:of\:moles\:of\:Solute}{Mass\:of\:solvent\:in\:kg}\\\\\\ \longrightarrow\sf m = \dfrac{24.5}{98 \times 0.4705}\\\\\\ \longrightarrow\sf m = \dfrac{0.25}{0.4705}\\\\\\ \longrightarrow{\boxed{\sf{\pmb{\pink{m = 0.531\:molal}}}}}\:\bigstar\\\\\\$

$\therefore$ Hence, The Molality of Given solution is 0.531 m.

250 cm3 of sulphur IC acid contain 24.5 g of h2so4. The density of solution is 1.98g/cm3. Determine molality

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