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3 years ago

How can a solution become saturated?

Chemistry

2 answers:

vladimir2022 [97]3 years ago

7 0

Answer:

it think it is more solute added

Liono4ka [1.6K]3 years ago

3 0

Explanation:

Answer :more solute is added

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4 years ago

A solution containing 3.90 g of an unknown nonelectrolyte liquid and 9.60 g water has a freezing point of −3.33 °C. The Kf = 1.8

pantera1 [17]

Answer:

The molar mass is 227 g/mol

Explanation:

Step 1: Data given

Mass of unknown nonelectrolyte = 3.90 grams

Mass of water = 9.60 grams

Freezing point of the solution = -3.33 °C

Kf = 1.86°C/m

Nonelectrolyte has a van't Hoff factor = 1

Step 2:

If you know the number of moles, and you know that is equivalent to 1.00 g, you can get molar mass.

∆T =i*m*K f

⇒ ∆T = difference in temperature between freezing point of solution and pure water = 3.33 °C

⇒ Van't hoff factor of the nonelectrolyte = 1

⇒ molality = moles nonelectrolyte / mass water

⇒ Kf = freezing point constant = 1.86 °C/m

3.33 = (1)(m)(1.86)

m = 1.79 molal = 1.79 moles / kg H2O

Step 3: Calculate moles nonelectrolyte

molality = moles / mass H2O

moles = molality * mass H2O

Moles = 1.79 molal * 0.0096

Moles = 0.017184‬ moles

Step 4: Calculate molar mass of nonelectrolyte

Molar mass = mass / moles

Molar mass = 3.90 grams / 0.017184 moles

Molar mass = 227 g/mol

The molar mass is 227 g/mol

5 0

3 years ago

A 3.00-L flask is filled with gaseous ammonia, NH3. The gas pressure measured at 27.0 ∘C is 2.55 atm . Assuming ideal gas behavi

Whitepunk [10]

Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.

Solution :

Using ideal gas equation,

$PV=nRT\\\\PV=\frac{w}{M}\times RT$

where,

n = number of moles of gas

w = mass of ammonia gas = ?

P = pressure of the ammonia gas = 2.55 atm

T = temperature of the ammonia gas = $27^oC=273+27=300K$

M = molar mass of ammonia gas = 17 g/mole

R = gas constant = 0.0821 L.atm/mole.K

V = volume of ammonia gas = 3.00 L

Now put all the given values in the above equation, we get the mass of ammonia gas.

$(2.55atm)\times (3.00L)=\frac{w}{17g/mole}\times (0.0821L.atm/mole.K)\times (300K)$

$w=5.28g$

Therefore, the mass of ammonia present in the flask in three significant figures are, 5.28 grams.

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4 years ago

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