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kifflom [539]
3 years ago
13

The vapor pressure of water at 45oC is 71.88 mmHg. What is the vapor pressure of a sugar (C12H22O11) solution made by dissolving

50.36 g of sugar in 88.69 g of water
Chemistry
1 answer:
Tju [1.3M]3 years ago
5 0

Answer:

69.79 mmHg is the pressure for the solution

Explanation:

Formula for vapor pressure lowering:

Vapor pressure of pure solvent(P°) - Vapor pressure of solution (P') = P° . Xm

Xm → Molar fraction of solute (moles of solute / Total moles)

Total moles = Moles of solute + Moles of solvent

Let's determine the moles:

50.36 g . 1mol/342 g = 0.147 moles of sugar

88.69 g. 1mol/ 18g = 4.93 moles of water

Total moles = 0.147 + 4.93 = 5.077 moles

Xm = 0.147 / 5.077 = 0.0289

If we replace data given in the formula:

71.88 mHg - P' = 71.88 mmHg . 0.0289 . 0.0289

P' = - (71.88 mmHg . 0.0289 - 71.88 mmHg)

P' = 69.79 mmHg

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<h3>Answer:</h3>

              Excess Reagent =  NBr₃

<h3>Solution:</h3>

The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,

                       2 NBr₃ + 3 NaOH   →   N₂ + 3 NaBr + 3 HBrO

Calculating the Limiting Reagent,

According to Balance equation,

               2 moles NBr₃ reacts with  =  3 moles of NaOH

So,

           40 moles of NBr₃ will react with  =  X moles of NaOH

Solving for X,

                       X  =  (40 mol × 3 mol) ÷ 2 mol

                       X  =  60 mol of NaOH

It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.

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Compound X has a molar mass of 416.48 g mol
bazaltina [42]

Answer:

P₂Cl₁₀

Explanation:

From the question given above, the following data were obtained:

Molar mass of compound X = 416.48 g/mol

Percentage of phosphorus (P) = 14.87%

Percentage of Chlorine (Cl) = 85.13%

Molecular formula of X =?

Next, we shall determine the empirical formula of compound X. This can be obtained as follow:

P = 14.87%

Cl = 85.13%

Divide by their molar mass

P = 14.87 / 31 = 0.480

Cl = 85.13 / 35.5 = 2.398

Divide by the smallest

P = 0.480 / 0.480 = 1

Cl = 2.398 / 0.480 = 5

Empirical formula of compound X is PCl₅

Finally, we shall determine the molecular formula of compound X. This can be obtained as follow:

Molar mass of compound X = 416.48 g/mol

Empirical formula = PCl₅

Molecular formula =?

Molecular formula= [Empirical formula]ₙ

[PCl₅]ₙ = 416.48

[31 + (35.5 × 5)]ₙ = 416.48

[31 + 177.5]n = 416.48

208.5n = 416.48

Divide both side by 208.5

n = 416.48 / 208.5

n = 2

Molecular formula = [PCl₅]ₙ

Molecular formula = [PCl₅]₂

Molecular formula = P₂Cl₁₀

Therefore, the molecular formula of compound X is P₂Cl₁₀

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