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frosja888 [35]
2 years ago
13

What is the mass of one mole of popcorn kernels? (sig fig hint: consider the precision of avogadro's number to determine the cor

rect number of sig figs.)?
Chemistry
2 answers:
sweet-ann [11.9K]2 years ago
7 0

Answer:

m=7.0106x10^22g

Explanation:

Hello,

In this case we must take into account that the average mass of a dozen popcorn kernels is 1.416 g, therefore, the average mass of one popcorn kernel is:

m=1.416g/12=0.118g

Now, we also know that one mole of popcorn kernels has 6.022x10²³ popcorn kernels, therefore the mass of one mole of popcorn kernels is computed as shown below:

m=1mol*\frac{6.022x10^{23}popcorn \ kernels }{1mol}*\frac{0.118g}{1popcorn\ kernels} \\m=7.0106x10^22g

Best regards.

svetlana [45]2 years ago
4 0

Since the mass of a single popcorn kernel is not given, lets us assume that it is 1 gram

Now:

1 mole of popcorn kernels contain 6.023 * 10^{23} kernels

Since:

1 kernel weighs 1 gram

Then,

1 mole = 6.023 *10^{23} kernels would weigh 6.023 * 10^{23} grams

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The given question is incomplete. The complete question is as follows.

When 70.4 g of benzamide (C_{7}H_{7}NO) are dissolved in 850 g of a certain mystery liquid X, the freezing point of the solution is 2.7^{o}C lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride (NH_{4}Cl) are dissolved in the same mass of X, the freezing point of the solution is 9.9^{o}C lower than the freezing point of pure X.

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    Moles of benzamide = \frac{mass}{\text{Molar mass of benzamide}}

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                    = 0.58 mol

Now, we will calculate the molality as follows.

     Molality = \frac{\text{moles of solute (benzamide)}}{\text{solvent mass in kg}}

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      dT = i \times K_{f} \times m,

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                  i = van't Hoff factor = 1 for non dissociable solutes

      K_{f} = freezing point constant of solvent

                m = 0.6837

Therefore, putting the given values into the above formula as follows.

             dT = i \times K_{f} \times m,

            2.7^{o}C = 1 \times K_{f} \times 0.6837 m

            K_{f} = 3.949 C/m

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 Moles of NH_{4}Cl = \frac{70.4 g}{53.491 g/mol}

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Hence, calculate the molality as follows.

    Molality = \frac{1.316 mol}{0.85 kg}

                  = 1.5484

It is given that value of change in temperature (dT) = 9.9^{o}C. Thus, calculate the value of Van't Hoff factor as follows.

              dT = i \times K_{f} \times m

   9.9^{o}C = i \times 3.949 C/m \times 1.5484 m

                     i = 1.62

Thus, we can conclude that the value of van't Hoff factor for ammonium chloride is 1.62.

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