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Given that the density of water is 0.975 g/mL and that 171 g of sucrose (molar mass: 342.30 g/mol) is dissolved in 512.85 mL of water at 80°C, what is the molality of this solution?

Respuesta :

josh09132
1.00 m is the molality of this solution

Answer: 1 molal

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

[tex]Molality=\frac{n\times 1000}{W_s}[/tex]

where,

n= moles of solute

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{171g}{342.30g/mol}=0.5moles[/tex]

 [tex]W_s[/tex] = weight of solvent in g

[tex]\text {Mass of solvent}={\text {density of solvent}\times {\text {Volume of solvent}=0.975g/ml\times 512.85ml=500g[/tex]

Putting in the values we get:

[tex]Molality=\frac{0.5\times 1000}{500}=1mole/kg[/tex]

Thus molality of solution will be 1 molal.

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