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PIT_PIT [208]
3 years ago
5

What is the boiling point of water when 175.0 g of Na2SO4, a strong electrolyte is dissolved in 1.000 Kg of water?

Chemistry
1 answer:
liubo4ka [24]3 years ago
8 0

Answer: 101.9^0C

Explanation:

Elevation in boiling point is given by:

\Delta T_b=i\times K_b\times m

\Delta T_b=T_b-T_b^0=(T_b-100)^0C = Elevation in boiling point

i= vant hoff factor = 3 (number of ions an electrolyte produce on complete dissociation)

Na_2SO_4\rightarrow 2Na^++SO_4^{2-}

K_f = freezing point constant = 0.512^0C/m

m= molality

\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (water)= 1.000 kg

Molar mass of solute Na_2SO_4 = 142 g/mol

Mass of solute Na_2SO_4  = 175.0 g

(T_b-100)^0C=3\times 0.512\times \frac{175.0g}{142g/mol\times 1.000kg}

T_b=101.9^0C

Thus the boiling point of water when 175.0 g of Na_2SO_4, a strong electrolyte is dissolved in 1.000 Kg of water is 101.9^0C

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