1) how many grams of copper(ii) sulfate pentahydrate (cuso4·5h2o) are needed to prepare 100.00 milliliters…

Explanation:

Step1: Calculate the number of moles

Use the formula $n = M\times V$, where $M$ is molarity and $V$ is volume in liters. First, convert volume to liters: $V=100.00\ mL = 0.10000\ L$, $M = 0.15\ mol/L$. $n=0.15\ mol/L\times0.10000\ L = 0.015\ mol$

Step2: Calculate the 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 the mass of solute

Use the formula $m = n\times M$. Here $n = 0.015\ mol$ and $M = 249.69\ g/mol$. $m=0.015\ mol\times249.69\ g/mol = 3.74535\ g\approx3.75\ g$

Answer:

$3.75$