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GuDViN [60]
3 years ago
5

Calculate the concentration, in moles per litre, of a solution formed when 13.0 g of calcium hydroxide is dissolved in 5.0 L. Pl

ease show work!
Chemistry
1 answer:
vichka [17]3 years ago
8 0

Answer: The concentration, in moles per litre, of a solution formed when 13.0 g of calcium hydroxide is dissolved in 5.0 L is 0.036 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

moles of Ca(OH)_2 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{13.0g}{74g/mol}=0.18mol

Now put all the given values in the formula of molality, we get

Molarity=\frac{0.18}{5.0L}=0.036M

Therefore, the concentration, in moles per litre, of a solution formed when 13.0 g of calcium hydroxide is dissolved in 5.0 L is 0.036 M

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How many moles of calcium carbonate-CaCO3 = 4.15 g​
marin [14]

Answer:

Number of moles = 0.042 mol

Explanation:

Given data:

Number of moles = ?

Mass of calcium carbonate = ?

Solution:

Formula:

Number of moles = mass/ molar mass

now we will calculate the molar mass of calcium carbonate.

atomic mass of Ca = 40 amu

atomic mass of C = 12 amu

atomic mass of O = 16 amu

CaCO₃ = 40 + 12+ 3×16

CaCO₃ = 40 + 12+48

CaCO₃ = 100 g/mol

Now we will calculate the number of moles.

Number of moles = 4.15 g / 100 g/mol

Number of moles = 0.042 mol

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Read 2 more answers
The vapor above a mixture of pentane and hexane at room temperature contains 35.5% pentane by mass. What is the mass percent com
FrozenT [24]
  • The mass percent of Pentane in solution is 16.49%
  • The mass percent of Hexane in solution is 83.51%

<u>Explanation</u>:

  • Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.
  • Convert these values to mol% using their molecular weights:

Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol

Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol

Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%

Hexane mol%: yh = 100 - 39.68 = 60.32%

Pp-vap = 425 torr = 0.555atm

Ph-vap = 151 torr = 0.199atm

  • From Raoult's law we know:  

Pp = xp \times Pp - vap = yp \times Pt                                       (1)

Ph = xh \times Ph - vap = yh \times Pt                                       (2)

  • Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:

(1 - xp) \times Ph - vap = (1 - yp) \times Pt                                   (3)

  • Substituting (1) into (3) we get:

(1-xp) \times Ph - vap = (1 - yp) \times xp \times Pp - vap / yp            (4)

  • Rearrange for xp:

xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap)      (5)

  • Subbing in the values we find:

Pentane mol% in solution: xp = 19.08%  

Hexane mol% in solution: xh = 80.92%

  • Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:

mp = 0.1908 mol \times 72.15 g/mol

     = 13.766 g

mh = 0.8092 mol \times 86.18 g/mol

     = 69.737 g

  • Mass% of Pentane solution = 13.766/(13.766+69.737)

                                                       = 16.49%

  • Mass% of Hexane solution  = 83.51%
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