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2 years ago

The mass of C02 contains same number of molecules as there are number of atoms in 2.3 g of sodium

Chemistry

1 answer:

Lunna [17]2 years ago

8 0

1- mole of any Substance will have the same number of Molecules.

Molecular weight of CO2= 44

Molecular weight of H20= 18

So 18 g of H20 will contain the same Number of molecules as in 44 g of CO2.

Therefore 7.2 g of H2O will have same Number of Moleculesas in =7.2/18 x 44

= 17.6 g of CO2.

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The boiling point of ethanol is 78.40 C. A student conducts an experiment and finds a boiling point of 75.70 C. What is the stud

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Answer:

3.40% corrected to 1dp

Explanation:

78.4-75.7=2.7

percentage error = error/original value× 100

= 2.7/78.40×100

= 3.44387755102040

= 3.40% corrected to 1dp

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3 years ago

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3 years ago

When 70.4 g of benzamide (C,H,NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2

Reil [10]

Answer:

1.60 is the van't Hoff factor for ammonium chloride in X.

Explanation:

$\Delta T_f=iK_f\times m$

$Delta T_f=K_f\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}$ ...(1)

where,

$\Delta T_f$ =Elevation in boiling point =

i = van't Hoff factor

$K_f$ = Freezing point constant

m = molality

1) When 70.4 g of benzamide are dissolved in 850. g of a certain mystery liquid X.

Mass of benzamide = 70.4 g

Molar mass of benzamide = 121 g/mol

i = 1 (organic molecule)

Mass of liquid X = 850 g = 0.850 kg

$K_f$ = Freezing point constant of liquid X= ?

$\Delta T_f=2.7^oC$

Putting all value in a (1):

$2.7^oC=K_f\times \frac{70.4 g}{121 g/mol\times 0.850 kg}$

$K_f=3.944 ^oC kg/mol$

2) When 70.4 g of ammonium chloride are dissolved in 850. g of a certain mystery liquid X.

Mass of ammonium chloride= 70.4 g

Molar mass of ammonium chloride = 53.5 g/mol

i = ? (ionic molecule)

Mass of liquid X = 850 g = 0.850 kg

$K_f=3.944 ^oC kg/mol$

$\Delta T_f=9.9^oC$

Putting all value in a (1):

$9.9^oC=i\times 3.944^oC kg/mol\times \frac{70.4 g}{53.5 g/mol\times 0.850 kg}$

i = 1.6011 ≈ 1.60

1.60 is the van't Hoff factor for ammonium chloride in X.

4 0

3 years ago