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Sliva [168]
3 years ago
15

Read the descriptions below of two substances and an experiment on each. Decide whether the result of the experiment tells you t

he substance is a pure substance or a mixture, if you can. • Sample A is a solid yellow cube with a total mass of 50.0 g. The cube is put into a beaker filled with 250. mL of water. The cube collapses into a small pile of orange powder at the bottom of the beaker. When this powder is filtered out, dried and weighed, it has a total mass of 29.9 g. If the experiment is repeated with 500. mL of water, the powder that's left over has a mass of 10.0 g. Sample B is 100. g of a coarse grey powder with a faint unpleasant smell. 15. mg of the powder are put into a very thin tube and heated. The powder begins melting at 66.2 °C.The temperature stays constant as the powder slowly melts. After the last of the powder melts, the temperature starts to rise again, eventually reaching 76.0 °C. O pure substance Is sample A made from a pure substance or a mixture? x ? mixture If the description of the substance and the outcome of the experiment isn't enough to decide, choose "can't decide. (can't decide) O pure substance Is sample B made from a pure substance or a mixture? If the description of the substance and the outcome of bstance and the outcome of the experiment isn't enough to decide, choose "can't decide." mixture (can't decide)
Chemistry
1 answer:
harkovskaia [24]3 years ago
7 0

Answer and Explanation: Sample A is a <u>mixture</u>. Solubility is characteristics of each substance, which means a substance can be distinguished from other substances and can be useful to separate mixtures.

In Sample A, when is added different volumes of water, the resulting powder has different mass. This means there are more than one substance forming the yellow cube. Therefore, sample A is a mixture.

Sample B is a <u>pure</u> <u>substance</u>. Each substance has its own melting point. Whe na pure substance reaches its melting point, temperature is constant until all of that substance is melted. In sample B, temperature is stable at 66.2°C and then, after all the powder is melted, it rises again. Therefore, sample B is a pure substance.

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Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6
Sergeeva-Olga [200]

Explanation:

The given data is as follows.

 Weight of solute = 75.8 g,   Molecular weight of solute (toulene) = 92.13 g/mol,    volume = 200 ml

  • Therefore, molarity of toulene is calculated as follows.

      Molarity = \frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}

                    = \frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}

                    = 4.11 M

Hence, molarity of toulene is 4.11 M.

  • As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

   Molality = \frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}

             = \frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}

             = 8.6 m

Hence, molality of given toulene solution is 8.6 m.

  • Now, calculate the number of moles of toulene as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{75.8 g}{92.13 g/mol}

                             = 0.8227 mol

Now, no. of moles of benzene will be as follows.

     No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{95.6 g}{78.11 g/mol}

                             = 1.2239 mol

Hence, the mole fraction of toulene is as follows.

         Mole fraction = \frac{\text{moles of toulene}}{\text{total moles}}

                             = \frac{0.8227 mol}{(0.8227 + 1.2239) mol}

                             = 0.402

Hence, mole fraction of toulene is 0.402.

  • As density of given solution is 0.857 g/cm^{3} so, we will calculate the mass of solution as follows.

         Density = \frac{mass}{volume}

     0.857 g/cm^{3} = \frac{mass}{200 ml}      (As 1 cm^{3} = 1 g)

                      mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

      Mass % = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100

                   = \frac{75.8 g}{171.4 g} \times 100

                   = 44.22%

Therefore, mass percent of toulene is 44.22%.

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