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Zielflug [23.3K]
4 years ago
13

An aqueous solution of glucose has molality of 6.81 and a density of 1.2 g/ml what is the molarity of glucose in the solution

Chemistry
1 answer:
andriy [413]4 years ago
3 0

Answer:

the molarity is 3.68 moles/L

Explanation:

the molality of the solution of sucrose is

m= moles of glucose / Kg of solvent (water)= 6.81 ,

since the molecular weight of glucose is 180.156 gr/mole , then per each kilogram of solvent there is

6.81 moles*180.156 gr/mole + 1000 gr of water = 2226.86 gr of solution

from the density

volume of solution = mass of solution/density = 2286.86 gr / 1.2 gr/ml = 1855.71 ml

therefore there is 1000 gr of water in 1855.71 ml

then the molarity M is

M= moles of glucose / L of solution = (moles of glucose /  Kg of solvent) *  (Kg of solvent/L of solution) = 6.81 moles/Kg * 1Kg/1.85 L = 3.68 moles/L

M= 3.68 moles/L

Note:

- Would be wrong in this case to assume density of water = 1 Kg/L since the solution is heavily concentrated in glucose and therefore the density of water deviates from its pure value.

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Name:_____________________________________________________ Date:___________ Period:_________ 3/23 - 3/27 Assignment 1: Gas Law P
crimeas [40]

Answer:

The answers are;

1. 8.2 liters

2. 1214.84 ml

3. 318.027 K

4. 4.00 l.

Explanation:

1. Boyle's law states that the volume of  given mass of gas is inversely proportional to its pressure at constant temperature

that is

P₁·V₁ = P₂·V₂

Where:

P₁ = Initial pressure = 40.0 mm Hg

V₁ = Initial volume = 12.3 liters

P₂ = Final pressure = 60.0 mm Hg

V₂ = Final volume = Required

From P₁·V₁ = P₂·V₂, V₂ is given by

V_2=\frac{P_1\cdot V_1}{P_2} = \frac{40.0 mm Hg\cdot 12.3 l}{60.0 mm Hg} =  8.2 l

The volume reduces to V₂ = 8.2 liters

2. Here Charles law states that

\frac{T_1}{V_1} =\frac{T_2}{V_2}

T₁ = Initial temperature = 27.0 °C = 300.15 K

V₁ = Initial volume = 900.0 mL

T₂ = Final temperature = 132.0 °C = 405.15 K

V₂ = Final volume = Required

Therefore  V_2 =\frac{T_2\cdot V_1}{T_1} = \frac{405.15 K\times 900.0 mL}{300.15 K} = 1214.84 ml

V₂ = 1214.84 ml

3.  Gay-Lussac's Law states that

\frac{T_1}{P_1} =\frac{T_2}{P_2}

Where:

P₁ = Initial pressure = 15.0 atmospheres

T₁ = Initial temperature = 25.0 °C = 298.15 K

P₂ = Final pressure = 16.0 atmospheres

T₂ = Final temperature = Required

∴ T_2 = \frac{T_1\times P_2}{P_1}

=  \frac{298.15 K\times 16.0atm}{15.0atm} = 318.027 K

T₂ = 318.027 K

4. Avogadro's law states that,

Equal volume of all gases at the same temperature and pressure contain equal number of molecules.

Therefore if 5.00 moles of gas occupies 2.00 l volume, then

1 moles will occupy 2.00/5 l volume and

10 moles will occupy 2.00/5 × 10 or 4.00 l volume.

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