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vlabodo [156]
4 years ago
7

What is the molarity of 12.5 g ch3oh in 50.0 ml of solution?

Chemistry
1 answer:
mr Goodwill [35]4 years ago
6 0
Data Given:
                   Mass  =  12.5 g

                   M.mass of Methanol  =  32.04 g/mol

                   Volume  =  50 ml ÷ 1000 =  0.05 L
Solution:

First of all calculate moles of Methanol,
  
                           Moles  =  Mass / M.mass

                           Moles  =  12.5 g / 32.04 g.mol⁻¹

                           Moles  =  0.39 mol

As we know Molarity is given as,

                          Molarity  =  Moles / Volume of Solution

Putting Values,

                          Molarity  =  0.39 mol ÷ 0.05 L

                          Molarity  =  7.80 mol.L⁻¹
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xxTIMURxx [149]

Answer:

25.08 grams of O₂ are needed to react with 8.15 g of C₂H₂.

Explanation:

The balanced reaction is:

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:

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Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

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Then you can apply the following rule of three: if by stoichiometry 52 grams of C₂H₂ react with 160 grams of O₂, 8.15 grams of C₂H₂ react with how much mass of O₂?

mass of O_{2} =\frac{8.15 grams of C_{2} H_{2}*160 grams of O_{2}  }{52 grams of C_{2} H_{2}}

mass of O₂= 25.08 grams

<u><em>25.08 grams of O₂ are needed to react with 8.15 g of C₂H₂.</em></u>

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