Question

1. how many grams of lithium sulfate are needed to make 250.0 ml of a 2.35 m solution?

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 the volume from mL to L: $V=250.0\ mL = 0.2500\ L$, and $M = 2.35\ M$. Then $n=2.35\ mol/L\times0.2500\ L = 0.5875\ mol$.

Step2: Determine the molar - mass of lithium sulfate

The formula of lithium sulfate is $Li_2SO_4$. The molar - mass of $Li$ is approximately $6.94\ g/mol$, the molar - mass of $S$ is approximately $32.07\ g/mol$, and the molar - mass of $O$ is approximately $16.00\ g/mol$. So, $M_{Li_2SO_4}=2\times6.94\ g/mol + 32.07\ g/mol+4\times16.00\ g/mol=109.94\ g/mol$.

Step3: Calculate the mass of lithium sulfate

Use the formula $m=n\times M$. Substitute $n = 0.5875\ mol$ and $M = 109.94\ g/mol$ into the formula: $m=0.5875\ mol\times109.94\ g/mol\approx64.6\ g$.

Answer:

$64.6\ g$