A non-electrolytic and non-volatile solute is added to pure water, difference between freezing point and boiling point is now 105°C. Calculate mass of solute present in 500 g of solvent. (Given molar mass of solute = 120 g/mol, kb = 0.512 °Cm–1, kf =1.86 °Cm–1)

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Vikas TU · Answer

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A non-electrolytic and non-volatile solute is added to pure water, difference between freezing point and boiling point is now 105°C. Calculate mass of solute present in 500 g of solvent. (Given molar mass of solute = 120 g/mol, kb = 0.512 °Cm–1, kf =1.86 °Cm–1)

A non-electrolytic and non-volatile solute is added to pure water, difference between freezing point and boiling point is now 105°C. Calculate mass of solute present in 500 g of solvent. (Given molar mass of solute = 120 g/mol, kb = 0.512 °Cm–1, kf =1.86 °Cm–1)

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A non-electrolytic and non-volatile solute is added to pure water, difference between freezing point and boiling point is now 105°C. Calculate mass of solute present in 500 g of solvent. (Given molar mass of solute = 120 g/mol, kb = 0.512 °Cm–1, kf =1.86 °Cm–1)

A non-electrolytic and non-volatile solute is added to pure water, difference between freezing point and boiling point is now 105°C. Calculate mass of solute present in 500 g of solvent. (Given molar mass of solute = 120 g/mol, kb = 0.512 °Cm–1, kf =1.86 °Cm–1)

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