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siniylev [52]
3 years ago
8

An aqueous solution of an unknown quantity of a nonelectrolyte solute is found to have a freezing point of −0.58°c.what is the m

olal concentration of the solution?
Chemistry
1 answer:
adell [148]3 years ago
4 0
Depression in freezing point is a colligative property i.e. it depends upon under  of solute particles present in solution.

Depression in freezing point (ΔTf) is mathematically expressed as,
ΔTf = Kf X m        ............... (1)
where Kf = cryoscopic constant
m = molality of solution

If solvent used is water, Kf = 1.853 K Kg/mol and Freezing point of pure solvent = 0 oC
∴ΔTf = 0.58 K         ............... (2)
Substituting 2 in 1 we get
0.58 = 1.853 X m
∴ m = 0.313 m

Thus, molality of solution is 0.313 m

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Calculate the number of moles of an ideal gas if it occupies 1750 dm3 under 125,000 Pa at a temperature of 127 C.
Phoenix [80]

Hey there!


* Converts 1750 dm³ in liters :


1 dm³ = 1 L so 1750 dm³ = 1750 liters



* Convertes 125,000 Pa in atm :


1 Pa = 9.86*10⁻⁶ atm so 9.86*10⁻⁶ / 125,000 => 1.233 atm


* Convertes 127ºC in K :


127 + 273.15 => 400.15 K


R = 0.082 atm.L/mol.K


Finally, it uses an equation of clapeyron :


p * V = n * R * T


1.233 * 1750 = n * 0.082 * 400.15


2157.75 = n * 32.8123


n = 2157.75 / 32.8123


n = 65.76 moles



hope this helps!



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Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "
Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

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Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

<em>As Mass = density × volume</em>

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

<em>PV = nRt</em>

<em>V = nRT/P</em>

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

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