Question #b6ba3

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Question #b6ba3

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Answer 1 · 2015-01-22T19:47:51

I'm assuming you mean $weight of$ $"weight of"$ , instead of $water of$ $"water of"$ , right? Here's how I think the data actually looks like:

The weight of the crucible is $10.0 g$ $"10.0 g"$ .
The weight of the crucible + the weight of the copper (II) sulfate pentahydrate is $14.98 g$ $"14.98 g"$ . Automatically, you know how much copper (II) sulfate pentahydrate you have

$m_{hydrate} = 14.98 g - 10.00 g = 4.980 g$ $m_("hydrate") = "14.98 g" - "10.00 g" = "4.980 g"$

The weight of the crucible + the weight of the residue is $13.54 g$ $"13.54 g"$ . The "residue" actually represents the anhydrated ${CuSO}_{4}$ $"CuSO"_4$ , which means that you now know how much water has been evaporated

$m_{water} = m_{hydrate} - (13.54 g - 10.00 g) = 4.980 g - 3.540 g$ $m_("water") = m_("hydrate") - "(13.54 g - 10.00 g)" = "4.980 g" - "3.540 g"$

$m_{water} = 1.440 g$ $m_("water") = "1.440 g"$

You know that water has a molar mass of $18.0 g/mol$ $"18.0 g/mol"$ . This will help you find the number of moles you have

$1.440 g \cdot \frac{1 mole}{18.0 g} = 0.0800$ $"1.440 g" * ("1 mole")/("18.0 g") = 0.0800$ $moles$ $"moles"$

The number of molecules is

$0.0800 moles \cdot \frac{6.022 \cdot 10^{23} molecules}{1 mole} = 4.82 \cdot 10^{22}$ $"0.0800 moles" * (6.022 * 10^(23) "molecules")/("1 mole") = 4.82 * 10^(22)$ $molecules$ $"molecules"$

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