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Gelneren [198K]
2 years ago
14

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?
Chemistry
2 answers:
nikitadnepr [17]2 years ago
7 0
1.00 m is the molality of this solution
ExtremeBDS [4]2 years ago
6 0

Answer: 1 molal

Explanation:

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

Molality=\frac{n\times 1000}{W_s}

where,

n= moles of solute

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

 W_s = weight of solvent in g

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

Putting in the values we get:

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

Thus molality of solution will be 1 molal.

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