Question

1. how many grams of copper(ii) sulfate pentahydrate (cuso₄·5 h₂o) are needed to prepare 100.00 milliliters of a 0.11 m copper(ii) sulfate solution?

Explanation:

Step1: Calculate moles of copper(II) sulfate

Use the formula $n = M\times V$, where $M$ is molarity and $V$ is volume in liters. Given $M = 0.11\ M$ and $V=100.00\ mL = 0.10000\ L$.
$n = 0.11\ mol/L\times0.10000\ L=0.011\ mol$

Step2: Determine molar mass of $CuSO_4\cdot5H_2O$

$Cu$: atomic mass = 63.55 g/mol, $S$: atomic mass = 32.07 g/mol, $O$: atomic mass = 16.00 g/mol, $H$: atomic mass = 1.01 g/mol.
$M_{CuSO_4\cdot5H_2O}=63.55 + 32.07+4\times16.00 + 5\times(2\times1.01+16.00)=249.69\ g/mol$

Step3: Calculate mass of $CuSO_4\cdot5H_2O$

Use the formula $m = n\times M$.
$m = 0.011\ mol\times249.69\ g/mol = 2.74659\ g\approx2.75\ g$

Answer:

$2.75$