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likoan [24]
4 years ago
12

Menthol is a crystalline substance with a peppermint taste and odor. When 1.48 g of menthol is dissolved in 25.0 g of cyclohexan

e, the freezing point of the solution is lowered by 7.90 °C. The freezing point and Kf constant for cyclohexane can be found here. Calculate the molar mass of menthol.
Chemistry
1 answer:
pishuonlain [190]4 years ago
4 0

Answer:

Molar mass of menthol is 155.86 g/mol

Explanation:

Let's apply the colligative property of freezing point depression:

ΔT = Kf . m . i

As menthol is an organic compound we assume the i (Van't Hoff factor as non electrolytic, so i = 1)

m = mol /kg (molality)

Kf = Cryoscopic constant, in this case, for cyclohexane

ΔT = T° freezing of pure solvent - T° freezing solution

7.90°C = 20.8°C / m . m . 1

7.90°C / 20.8 m/°C = m

0.380 m = molality

0.380 moles are the moles of solute (menthol) dissolved in 1kg of solvent, but our mass of solvent is 25 g.

Let's convert the g of solvent to kg to work with molality

25 g = 0.025 kg

molal . solvent mass = moles of solute

0.380 mol /kg . 0.025kg = 9.49×10⁻³ moles

Now, we have the moles and the mass of solute. Let's determine the molar mass of menthol (grams / moles)

1.48 g / 9.49×10⁻³ m = 155.86 g/m

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1. A gas has a volume of 45 L at a pressure of 140 kPa. What is the volume when
pishuonlain [190]

Answer:

The new volume of the gas is 21 L.

Explanation:

Volume of a gas is inversely proportional to its pressure at constant temperature such that,

V\propto \dfrac{1}{P}

or

P_1V_1=P_2V_2

We have,

V_1=45\ L\\\\P_1=140\ kPa \\\\P_2=300\ kPa

It is required to find V₂. Using above law or Boyle's law such that :

V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{140\times 45}{300}\\\\V_2=21\ L

So, the new volume of the gas is 21 L.

7 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.

8 0
3 years ago
A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur
dsp73

Answer:

12.8 g of O_{2} must be withdrawn from tank

Explanation:

Let's assume O_{2} gas inside tank behaves ideally.

According to ideal gas equation- PV=nRT

where P is pressure of O_{2}, V is volume of O_{2}, n is number of moles of O_{2}, R is gas constant and T is temperature in kelvin scale.

We can also write, \frac{V}{RT}=\frac{n}{P}

Here V, T and R are constants.

So, \frac{n}{P} ratio will also be constant before and after removal of O_{2} from tank

Hence, \frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}

Here, \frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm} and P_{after}=17atm

So, n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol

So, moles of O_{2} must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of O_{2} = 32 g/mol

So, mass of O_{2} must be withdrawn = (32\times 0.40)g=12.8g

7 0
3 years ago
2. What is an example of a molecule<br> with different types of atoms?
agasfer [191]

Answer:

a compound

Explanation:

A molecule with multiple atoms is compound like water it's made up of 1 oxygen and two hydrogen.

4 0
3 years ago
Chemist question, helppp pls
Doss [256]

1: it is +2

2: it is +6

(Make this brainliest answer please)

8 0
3 years ago
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