1. Define the following terms: 1. Mole fraction. 2. Van’t Hoff factor. 2. 100 mg of a protein is dissolved in enough water to make 10.0 mL of a solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25^ C, what is the molar mass of protein? (R = 0.0821 L atm mol^ –1 K^ –1 and 760 mm Hg = 1 atm.)

1. 1. Mole fraction is the ratio of number of moles of one component to the total number of moles in a mixture. 2. Van't Hoff factor is expressed as: i= norml molar mass (m_ 1 ) abnormal molar mass (m'_ 1 )(due to dissociation or association ) 2. = CRT M_ 2 = w2RT V M_ 2 = 100 10^ -3 g 0.0821 L atm mol^ -1 K^ -1 298 K 760 1000 13.3 atm 10L M_2 = 13980 g mol^ –1 or 1.4 10^4 g mol^ –1 .

1. Define the following terms: 1. Mole fraction. 2. Van’t Hoff fa…

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Question

Define the following terms:

Mole fraction.
Van’t Hoff factor.

$100\ mg$ of a protein is dissolved in enough water to make $10.0\ mL$ of a solution. If this solution has an osmotic pressure of $13.3\ mm\ Hg$ at $25^\circ C,$ what is the molar mass of protein$?$

$(R = 0.0821\ L\ atm\ mol^{–1} K^{–1}$ and $760\ mm\ Hg = 1\ atm.)$

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Answer

Mole fraction is the ratio of number of moles of one component to the total number of moles in a mixture.
Van't Hoff factor is expressed as:

$\text{i}=\frac{\text{norml molar mass (m}_{1})}{\text{abnormal molar mass (m'}_{1})\text{(due to dissociation or association )}}$

$\pi = CRT$

$\text{M}_{2}=\frac{\text{w2RT}}{\pi\text{ V}}$
$\text{M}_{2}=\frac{100\times10^{-3}\text{g}\times0.0821\text{ L atm mol}^{-1}\text{K}^{-1}\times\text{298 K}\times760\times1000}{\text{13.3 atm}\times\text{10L}}$
$M_2 = 13980\ g\ mol^{–1}$ or $1.4 \times 10^4\ g \ mol^{–1}.$

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