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madam [21]
2 years ago
7

Dissolve 30 g of sodium sulphate into 300 mL of water

Chemistry
1 answer:
Aneli [31]2 years ago
3 0

Answer:

number of moles = 0.21120811

Explanation:

To find the number of moles, given the mass of the solute, we use the formula:

\mathrm{n =   \dfrac{ m  }{ M  } }

\mathrm{n = number\:of\:moles\:(mol)}

\mathrm{m = mass\:of\:solute\:(g)}

\mathrm{M = molar\:mass\:of\:solute\:(  \dfrac{ g  }{ mol  }   )}

Label the variables with the numbers in the problem:

\mathrm{n =\:?}

\mathrm{m =30\:g }

\mathrm{M =\:?\:Calculate\:the\:molar\:mass }

The first thing we have to do is find the molar mass of sodium sulfate, in order for us to use the formula for finding the number of moles:

Formula for finding the molar mass of sodium sulfate:

M({ \left Na \right }_{ 2  }   { \left So \right }_{ 4  })   =  m \left( Na  \right)  +m \left( S  \right)  +m \left( O  \right)

For the variables and what they mean are below for finding the molar mass of sodium sulfate:

\mathrm{M =molar\:mass }

\mathrm{m =moles=2\:moles\:for\:Na\:,1\:mole\:for\:S,\:and\:4\:moles\:for\:O}

\mathrm{Na =sodium=22.99\:g }

\mathrm{S =sulfur=32.06\:g }

\mathrm{O =oxygen=16.00\:g }

Plug the numbers into the formula, to find the molar mass of sodium sulfate:

M({ \left Na \right }_{ 2  }   { \left So \right }_{ 4  })   =  m \left( Na  \right)  +m \left( S  \right)  +m \left( O  \right)

\mathrm{Substitute\:the\:values\:into\:the\:formula}

M  =  2 \left( 22.99  \right)  +1 \left( 32.06  \right)  +4 \left( 16.00  \right)

\mathrm{Multiply\:2\:by\:22.99\:to\:get\:45.98\:and\:1\:by\:32.06\:to\:get\:32.06}

\mathrm{M =  45.98+32.06+4\:(16)}

\mathrm{Multiply\:4\:by\:16\:to\:get\:64}

\mathrm{M =  45.98+32.06+64}

\mathrm{Add\:45.98\:and\:32.06\:to\:get\:78.04}

\mathrm{M =  78.04+64}

\mathrm{Add\:78.04\:and\:64\:to\:get\:142.04}

\mathrm{M =  142.04}

Now that we have found the molar mass, we can calculate the number of moles in the solution of sodium sulfate with the formula:

\mathrm{n =   \dfrac{ m  }{ M  } }

\mathrm{n =\:?}

\mathrm{m =30\:g }

\mathrm{M = 142.04\:g/mol}

\mathrm{Substitute\:the\:values\:into\:the\:formula}

\mathrm{n =   \dfrac{ 30  }{ 142.04  }}

\mathrm{Divide\:142.04\:by\:30\:to\:get\:0.21120811}

\mathrm{n =  0.21120811}

0.21120811 rounded gives you 0.2112

or if you did the problem without decimals

30 grams of sodium sulfate divided by its molecular weight – which we found to be 142 – gives us a value of 0.2113 moles.

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C. If 62.9 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were
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Explanation:

Given data:

Mass of lead(II) chloride produced = 62.9 g

Mass of lead(II) nitrate used = ?

Solution:

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Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:

                            PbCl₂        :          Pb(NO₃)₂

                               1             :             1

                             0.23         :            0.23

Mass of Pb(NO₃)₂:

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Mass = 76.176 g

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