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erik [133]
1 year ago
14

A 125-mL sample of an aqueous solution of the protein ovalbumin from chicken egg white contains 1.31 g of the dissolved protein

and has an osmotic pressure of 4.32 torr at 25°C. What is the molar mass of ovalbumin?
Chemistry
1 answer:
Stella [2.4K]1 year ago
7 0

The molar mass of the protein is 45095 g/mol.

The mass of a sample of a chemical compound divided by the quantity, or number of moles in the sample, measured in moles, is known as the molar mass of that compound.

The expression of molar mass of protein is

M₂ = (W₂/P) (RT/V)

Given;

W₂ = 1.31g

P = 4.32 torr = 5.75 X 10⁻³ bar

R = 0.083 Lbar/mol/K

T = 25°C = 298.15 K

V = 125 ml = 0.125 L

Putting all the values in the above formula

M₂= (1.31 g/5.75 X 10⁻³ bar) X (0.083 Lbar/mol/K X 2)/0.125 L)

M₂ = 45095 g/mol

Thus, the molar mass of the protein is 45095 g/mol.

Learn more about the Molar mass with the help of the given link:

brainly.com/question/22997914

#SPJ4

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Read 2 more answers
A solution is prepared by adding 50.00 g of lactose (milk sugar) to 110.0 g of water at 55 °C. The partial pressure of water abo
Ede4ka [16]

Answer:

\boxed{\text{115.2 torr}}

Explanation:

Let’s call water Component 1 and lactose Component 2.

According to Raoult’s Law,  

p_{1} = \chi_{1}p_{1}^{\circ} \text{ and}\\p_{2} = \chi_{2}p_{2}^{\circ}

where

p₁ and p₂ are the vapour pressures of the components above the solution

χ₁ and χ₂ are the mole fractions of the components

p₁° and p₂° are the vapour pressures of the pure components.

Data:

m₁ = 110.0 g;    p₁° = 118.0 torr

m₂ = 50.00 g; p₂° =    0    torr

1. Calculate the moles of each component

n_{1} = \text{110.0 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{6.104 mol}\\\\n_{2} = \text{50.00 g} \times \dfrac{\text{1 mol}}{\text{342.3 g}} = \text{0.1461 mol}

2. Calculate the mole fraction of each component

\begin{array}{rcl}\chi_{2} & = & \dfrac{n_{2}}{n_{1} + n_{2}}\\\\&= & \dfrac{0.1461}{6.104 + 0.1461}\\\\& = &\dfrac{0.1461}{6.250}\\\\& = & \mathbf{0.023 37}\\\chi_{1}& = &1 - \chi_{2}\\& = &1 - 0.023 37\\& = & \mathbf{0.976 63}\\\end{array}

3. Calculate the vapour pressure of the mixture

p_{1} = 0.97663 \times \text{118.0 torr} = \text{ 115.2 torr}\\p_{2} = 0\\p_{\text{tot}} = p_{1} + p_{2} = \text{115.2 torr + 0} = \textbf{115.2 torr}\\\text{The partial pressure of water above the solution is $\boxed{\textbf{115.2 torr}}$}

8 0
3 years ago
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