
What is the amount (in moles) of copper (II) sulfate pentahydrate that you will need to make 100mL of a 0.500 mol/L solution, with all the calculations?
Answer
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Hint: Copper (II) sulfate pentahydrate has the formula, $CuS{{O}_{4}}.5{{H}_{2}}O$, it contain water of hydration which are 5. Number of moles of any substance is the given mass upon the molecular mass of that substance. For calculating the number of moles of copper (II) sulfate, keep in mind the water of hydration.
Formula used: $n=C.V$ , where n is number of moles of solute, C is concentration and V is volume.
Number of moles = $\dfrac{mass}{molar\,mass}$
Complete answer:
We have to find the amount of copper (II) sulfate pentahydrate in moles, which will be needed to make 100mL of a 0.500 mol/L solution. For this we will first take out the number of moles of copper (II) sulfate, and then calculate the molar mass of the whole hydrate salt. Then using the number of moles expression, the amount in the calculated moles will be determined.
As numbers of moles are the concentration multiplied with volume, we have, $n=C.V$, so
n = 0.500 $\times $ 0.1
n = 0.05 moles
Now, we have 0.05 moles of $CuS{{O}_{4}}.5{{H}_{2}}O$.
The molar mass of $CuS{{O}_{4}}.5{{H}_{2}}O$is calculated as, mass of $\left( Cu+S+4\times O \right)+\left( 5\times {{H}_{2}}O \right)$ , so putting the respective masses as Cu (63.5), S (32), O (16), and water (18), we have
$\left( 63.5+32+4\times 16 \right)+\left( 5\times 18 \right)$= 249.5 g/mol
So, the molecular mass of $CuS{{O}_{4}}.5{{H}_{2}}O$is 249.5 g/mol, now using number of moles and mass relationship, we will calculate the mass of$CuS{{O}_{4}}.5{{H}_{2}}O$
Number of moles = $\dfrac{mass}{molar\,mass}$
Mass = number of moles $\times $molar mass
Mass = 0.05 moles $\times $249.5 g/mol
Mass = 12.48 g
Hence, 0.05 moles per 12.48 g of copper (II) sulfate pentahydrate is needed to make a solution of 100mL of 0.500 mol/L.
Note:
For making this solution, dissolve 12.48 g of the solid in some water, and then make the volume to 100mL. As water of hydration, has 4.5 g amount, so add up to 95mL of water to reach a total 100mL. the mass of water of hydration is calculated as, for every 1 mole of $CuS{{O}_{4}}$, 5 moles of ${{H}_{2}}O$ are needed, so moles of ${{H}_{2}}O$ will be, $5\times 0.05$ = 0.25 moles. These are then multiplied by molar mass of water to get 4.5 g.
Formula used: $n=C.V$ , where n is number of moles of solute, C is concentration and V is volume.
Number of moles = $\dfrac{mass}{molar\,mass}$
Complete answer:
We have to find the amount of copper (II) sulfate pentahydrate in moles, which will be needed to make 100mL of a 0.500 mol/L solution. For this we will first take out the number of moles of copper (II) sulfate, and then calculate the molar mass of the whole hydrate salt. Then using the number of moles expression, the amount in the calculated moles will be determined.
As numbers of moles are the concentration multiplied with volume, we have, $n=C.V$, so
n = 0.500 $\times $ 0.1
n = 0.05 moles
Now, we have 0.05 moles of $CuS{{O}_{4}}.5{{H}_{2}}O$.
The molar mass of $CuS{{O}_{4}}.5{{H}_{2}}O$is calculated as, mass of $\left( Cu+S+4\times O \right)+\left( 5\times {{H}_{2}}O \right)$ , so putting the respective masses as Cu (63.5), S (32), O (16), and water (18), we have
$\left( 63.5+32+4\times 16 \right)+\left( 5\times 18 \right)$= 249.5 g/mol
So, the molecular mass of $CuS{{O}_{4}}.5{{H}_{2}}O$is 249.5 g/mol, now using number of moles and mass relationship, we will calculate the mass of$CuS{{O}_{4}}.5{{H}_{2}}O$
Number of moles = $\dfrac{mass}{molar\,mass}$
Mass = number of moles $\times $molar mass
Mass = 0.05 moles $\times $249.5 g/mol
Mass = 12.48 g
Hence, 0.05 moles per 12.48 g of copper (II) sulfate pentahydrate is needed to make a solution of 100mL of 0.500 mol/L.
Note:
For making this solution, dissolve 12.48 g of the solid in some water, and then make the volume to 100mL. As water of hydration, has 4.5 g amount, so add up to 95mL of water to reach a total 100mL. the mass of water of hydration is calculated as, for every 1 mole of $CuS{{O}_{4}}$, 5 moles of ${{H}_{2}}O$ are needed, so moles of ${{H}_{2}}O$ will be, $5\times 0.05$ = 0.25 moles. These are then multiplied by molar mass of water to get 4.5 g.
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