What is the molality of a solution when 0.50 moles glucose is dissolved in 750 g of water?

Chemistry

1 answer:

xxMikexx [17]2 years ago

3 0

Answer:

0.67mol/Kg

Explanation:

The following were obtained from the question:

Mole of solute = 0.50mol

Mass of solvent = 750g = 750/1000 = 0.75Kg

Molality =?

Molality = mole of solute /mass of solvent

Molality = 0.5/0.75

Molality = 0.67mol/Kg

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Antarctica, almost completely covered in ice, has an area

Marysya12 [62]

<u>Answer:</u> The mass of ice is $2.39\times 10^{22}g$

<u>Explanation:</u>

We are given:

Area of Antarctica = $5,500,000mi^2=5,500,000\times 2.59\times 10^{10}=142.45\times 10^{15}cm^2$ (Conversion factor: $1mi^2=2.59\times 10^{10}cm^2$ )

Height of Antarctica with ice = 7500 ft.

Height of Antarctica without ice = 1500 ft.

Height of ice = 7500 - 1500 = 6000 ft = $182.88\times 10^3cm$ (Conversion factor: 1 ft = 30.48 cm)

To calculate mass of ice, we use the equation:

$\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}$

We are given:

Density of ice = $0.917g/cm^3$

Volume of ice = Area × Height of ice = $142.45\times 10^{15}cm^2\times 182.88\times 10^3cm=26051.26\times 10^{18}cm^3$

Putting values in above equation, we get:

$0.917g/cm^3=\frac{\text{Mass of ice}}{26051.26\times 10^{18}cm^3}\\\\\text{Mass of ice}=(0.917g/cm^3\times 26051.26\times 10^{18}cm^3=2.39\times 10^{22}g$