Question

The weight of $1 \times 10^{22}$ molecules of $CuSO_4.5H_2O$ is ...................

Answer 1

4.144 g

Answer 2

To find the weight of $1 \times 10^{22}$ molecules of $CuSO_4.5H_2O$, we need to follow these steps:

1. Calculate the molar mass of $CuSO_4.5H_2O$:
   We use the standard atomic masses:
   Cu = 63.5 g/mol
   S = 32.0 g/mol
   O = 16.0 g/mol
   H = 1.0 g/mol

   Molar mass of $CuSO_4$ = M(Cu) + M(S) + 4 * M(O) = 63.5 + 32.0 + 4 * 16.0 = 63.5 + 32.0 + 64.0 = 159.5 g/mol.
   Molar mass of $H_2O$ = 2 * M(H) + M(O) = 2 * 1.0 + 16.0 = 2.0 + 16.0 = 18.0 g/mol.
   Molar mass of $5H_2O$ = 5 * M($H_2O$) = 5 * 18.0 = 90.0 g/mol.

   Molar mass of $CuSO_4.5H_2O$ = M($CuSO_4$) + M($5H_2O$) = 159.5 + 90.0 = 249.5 g/mol.

2. Determine the number of moles:
   We are given the number of molecules as $1 \times 10^{22}$.
   Avogadro's number ($N_A$) is $6.022 \times 10^{23}$ molecules/mol.

   Number of moles = (Number of molecules) / ($N_A$)
   Number of moles = $(1 \times 10^{22}) / (6.022 \times 10^{23})$ mol
   Number of moles = $(1 / 6.022) \times 10^{(22-23)}$ mol
   Number of moles = $(1 / 6.022) \times 10^{-1}$ mol

3. Calculate the weight (mass):
   Weight = Number of moles * Molar mass
   Weight = $((1 / 6.022) \times 10^{-1} \text{ mol}) \times (249.5 \text{ g/mol})$
   Weight = $(249.5 / 6.022) \times 10^{-1}$ g

   Let's calculate the value:
   $249.5 / 6.022 \approx 41.4397$

   Weight $\approx 41.4397 \times 10^{-1}$ g
   Weight $\approx 4.14397$ g

Given the number of molecules $1 \times 10^{22}$, if interpreted as having 2 significant figures ($1.0 \times 10^{22}$), and using $N_A$ with 4 significant figures ($6.022 \times 10^{23}$) and Molar Mass with 4 significant figures (249.5), the result should be reported with 2 significant figures, which would be 4.1 g. However, in such problems, it is common practice to use the standard constants and report the result with a precision consistent with those constants (usually 3-4 significant figures) unless the input number clearly limits the precision to fewer significant figures. Assuming at least 3 significant figures are expected.

Using the calculated value $\approx 4.14397$ g.
Rounding to 3 significant figures gives 4.14 g.
Rounding to 4 significant figures gives 4.144 g.

Let's provide the answer with 3 decimal places for sufficient precision.

The weight is approximately 4.144 g.