# Answer to Question #56160 in General Chemistry for sue

Question #56160

A 2.50 g sugar cube (sucrose: C12H22O11) is dissolved in a 350 mL teapot containing 80∘ C water (density of water at 80∘ C= 0.975 g/mL). What is the molality of the sugar solution?

Express the molality with the appropriate units.

I entered these and they are not correct, can you help?

ANSWER 1: Deduction: -3%

0.03molkg

ANSWER 2: Deduction: -3%

343.77gmol

Enter your answer using dimensions of concentration.

Express the molality with the appropriate units.

I entered these and they are not correct, can you help?

ANSWER 1: Deduction: -3%

0.03molkg

ANSWER 2: Deduction: -3%

343.77gmol

Enter your answer using dimensions of concentration.

Expert's answer

b = n (of solute) / m (of solvent)

solvent – H

solute – C

ρ (density) = m / V (mass / volume),

so m = ρ * V

m (H

n = m / M

n (C

b (solution) = 0.0073 mol / 0.3413 g = 0.021 mol/kg

Answer: 0.021 mol/kg

solvent – H

_{2}Osolute – C

_{12}H_{22}O_{11}ρ (density) = m / V (mass / volume),

so m = ρ * V

m (H

_{2}O) = 0.975 (g/mL) * 350 mL = 341.25 gn = m / M

n (C

_{12}H_{22}O_{11}) = 2.5 g / 342 g/mol = 0.0073 molb (solution) = 0.0073 mol / 0.3413 g = 0.021 mol/kg

Answer: 0.021 mol/kg

## Comments

## Leave a comment