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Ainat [17]
3 years ago
12

A certain substance X has a normal boiling point of 117.8°C and a molal boiling point elevation constant =Kb·0.63°C·kgmol−1. A s

olution is prepared by dissolving some urea NH22CO in 900.g of X. This solution boils at 118.3°C. Calculate the mass of urea that was dissolved. Round your answer to 1 significant digit.
Chemistry
1 answer:
OlgaM077 [116]3 years ago
5 0

Answer:

42.8 g was the mass of dissolved urea.

Explanation:

Boiling point elevation to solve this:

ΔT = Kb . m

where ΔT → Boiling T° of solution - Boiling T° of pure solvent

Let's replace the data given:

118.3°C - 117.8°C = 0.63°C/m . m

0.5°C / 0.63 m/°C = m → 0.793 mol/kg

Mass of solvent → 900 g → g to kg → 900 g . 1kg/1000 = 0.9 kg

Molality . kg = moles → 0.793 mol/kg . 0.9 kg = 0.714 mol

These are the moles of urea we used. Let's determine the mass dissolved.

0.714 mol . 60 g /1mol = 42.8 g

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8 0
4 years ago
What is the percent by mass of oxygen in carbon dioxide
san4es73 [151]
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6 0
3 years ago
How would you prepare 2.5 L of a 0.800M solution of KNO3?
statuscvo [17]
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Molar mass KNO</span>₃ = <span>101.1032 g/mol

mass =  Molarity * molar mass * volume

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mass = 202.2064 g of KNO</span>₃

<span>To prepare 2.5 L (0800 M) of KNO3 solution, must weigh 202.2064 g of salt, dissolve in a Beker, transfer with the help of a funnel of transfer to a volumetric flask, complete with water up to the mark, capping the balloon and finally shake the solution to mix.</span>

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7 0
3 years ago
How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?
shepuryov [24]

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically,         molarity = \frac{no. of moles}{Volume (in L) of solution}

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

           molarity = \frac{no. of moles}{Volume of solution in liter}

            0.0800 M = \frac{no. of moles}{0.05 L}

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Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

                No. of moles = \frac{mass in grams}{molar mass}

             mass in grams = no. of moles \times molar mass of CuSO_{4}.5H_{2}O

                                       = 1.6 mol \times 249.68 g/mol

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Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

4 0
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