SATHEE: Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry - Result Question 102

####40. The mole fraction of a solute in a solution is 0.1 . At $298 K$ , molarity of this solution is the same as its molality. Density of this solution at $298 K$ is $2.0 g c m^{- 3}$ . The ratio of the molecular weights of the solute and solvent, $(\frac{m_{solute}}{m_{solvent}})$ is …

(2016 Adv.)

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Solution:

Moles of solute, $n_{1} = \frac{w_{1}}{m_{1}}$ ; Moles of solvent, $n_{2} = \frac{w_{2}}{m_{2}}$

$χ_{1} (solute) = 0.1 and χ_{2} (solvent) = 0.9$

$∴ \frac{χ_{1}}{χ_{2}} = \frac{n_{1}}{n_{2}} = \frac{w_{1}}{m_{1}} \cdot \frac{m_{2}}{w_{2}} = \frac{1}{9}$

Molarity $= \frac{Solute (moles)}{Volume (L)} = \frac{w_{1} \times 1000 \times 2}{m_{1} (w_{1} + w_{2})}$

Note Volume $= \frac{Total mass of solution}{Density} = (\frac{w_{1} + w_{2}}{2}) m L$

Molality $= \frac{Solute (moles)}{Solvent (k g)} = \frac{w_{1} \times 1000}{m_{1} \times w_{2}}$

Given,

molarity $=$ molality

hence,

$\frac{2000 w_{1}}{m_{1} (w_{1} + w_{2})} = \frac{1000 w_{1}}{m_{1} w_{2}}$

$∴ \frac{w_{2}}{w_{1} + w_{2}} = \frac{1}{2} \Rightarrow w_{1} = w_{2} = 1$

$∴ \frac{w_{1} m_{2}}{m_{1} w_{2}} = \frac{1}{9} \Rightarrow \frac{m_{1} (solute)}{m_{2} (solvent)} = 9$

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Some Basic Concepts of Chemistry - Result Question 102

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Some Basic Concepts of Chemistry - Result Question 102

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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