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What mass of Na2SO4 must be dissolved in 104 grams of water to lower the freezing point by 2.50 °C? The freezing point depression constant, Kfp, of water is –1.86 °C/m. Assume the van't Hoff factor for Na2SO4 is 2.88.

Respuesta :

Answer: The mass of [tex]Na_2SO_4[/tex] that must be dissolved is 6.89 grams.

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=2.50^0C[/tex] = Depression in freezing point

i= vant hoff factor = 2.88

[tex]K_f[/tex] = freezing point constant = [tex]1.86^0C/m[/tex]

m= molality

[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

Weight of solvent (water)= 104 g = 0.104 kg  

Molar mass of [tex]Na_2SO_4[/tex] = 142 g/mol

Mass of [tex]Na_2SO_4[/tex] added = ?

[tex]2.50=2.88\times 1.86\times \frac{x}{142\times 0.104}[/tex]

[tex]x=6.89g[/tex]

The mass of [tex]Na_2SO_4[/tex] that must be dissolved is 6.89 grams.

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