Chemistry, 18.04.2020 00:56 lluvihorsechief01

What is the molality of a solution containing 100 grams of glucose (C6H12O6, molar mass = 180 g/mol) dissolved in 2.5 kg of water?

0.22 m

72 m

7200 m

40 m

Answers: 3

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Answers

Answer from: lyn94

The molality = 0.22 m (option 1 is correct)

Explanation:

Step 1: Data given

Mass of glucose = 100 grams

Molar mass glucose = 180 g/mol

Mass of water = 2.5 kg

Step 2: Calculate moles glucose

Moles glucose = mass glucose / moalr mass glucose

Moles glucose = 100 grams / 180 g/mol

Moles glucose = 0.556 moles

Step 3: Calculate molality

Molality = moles glucose / mass water

Molality = 0.556 moles / 2.5 kg

Molality = 0.2224 molal

The molality = 0.22 m (option 1 is correct)

Answer from: Quest

when the volume of gas is changed from 717.73 ml to 852 ml the temperature will change from 315 c to 452 c

calculation

this is calculated using the charles law formula

that is, v1/t1 = v2/t2

v1=?

t1 = 315 c into kelvin = 315 +273 = 588 k

v2 = 852 ml

t2 = 425c + 273 =698 k

make v1 the subject of the formula ( by multiplying both side by t1)

v1=( v2 xt1) / t2

v1= (852 ml x 588 k)/ 698 k= 717.73 ml

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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What is the molality of a solution containing 100 grams of glucose (C6H12O6, molar mass = 180 g/mol)...