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podryga [215]
3 years ago
9

The molar masses of three gases, A, B, and C, are 4 grams, 16 grams, and 32 grams respectively. Which sentence represents the ef

fusion rates of these gases?
Chemistry
2 answers:
Colt1911 [192]3 years ago
7 0

B for plato users C<B<A

timurjin [86]3 years ago
3 0
This is where something called Graham's Law applies. Grahams law says that the rate of effusion is inversely proportional to the square root of the molar mass of a gas. What that means is that the less the molar mass of a gas is, the less quickly the gas effuses (effusion being the rate at which a gas can travel through a small hole). A great way to think about this is that effusion is thought about as though the gas is traveling through a small hole, so smaller particles would be able to go through it with greater ease than would a large particle. I don't know what particular sentences the question asks for are, but the answer should be that gas A (molar mass 4) has the greatest effusion rate, gas B (molar mass 16) has the second fastest effusion rate, and gas C (molar mass 32) has the slowest effusion rate.
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The given question is incomplete. The complete question is as follows.

A certain liquid has a normal boiling point of 124.2^{o}C and a boiling point elevation constant k_{b} = 0.62 ^{o}C kg mol^{-1}. A solution is prepared by dissolving some sodium chloride (NaCl) in 6.50 g of X. This solution boils at 127.4^{o}C. Calculate the mass of NaCl that was dissolved. Round your answer to significant digits.

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As per the colligative property, the elevation in boiling point will be as follows.

     

T = boiling point of the solution =

T_{o} = boiling point of the pure solvent = 124.2^{o}C

K_{b} = elevation of boiling constant = 0.62 ^{o}C kg mol^{-1}

We will calculate the molality as follows.

     molality = \frac{\text{maas of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in g}}

i = vant hoff's factor

As NaCl is soluble in water and dissociates into sodium and chlorine ions so i = 2.

Putting the given values into the above formula as follows.  

    T - T_{o} = i \times K_{b} \times \text{molality of solution}

  (127.4 - 124.2)^{o}C = 2 \times 0.62 \times \frac{m}{60} \times \frac{1000}{650}

                  m = 100 g

Therefore, we can conclude that 100 g of NaCl was dissolved.

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