Answer:
a. 7,24g of CsBr
b. 4,63g of CaSO₄
c. 5,57g of Na₃PO₄
d. 7,82g of Li₂Cr₂O₇
e. 5,65g of K₂C₂O₄
Explanation:
To make 3,40x10² mL of a 0,100M solution you need:
3,40x10² mL × 0,100 mmol/mL = 34mmol ≡ 0,034moles of solute.
a. 0,034 moles of CsBr are:
0,034 mol CsBr× = <em>7,24g of CsBr </em><em>-Where 212,81g/mol is molar mass of CsBr-</em>
b. 0,034 moles of CaSO₄ are:
0,034 mol CaSO₄× = <em>4,63g of CaSO₄</em>
c. 0,034 moles of Na₃PO₄ are:
0,034 mol Na₃PO₄× = <em>5,57g of Na₃PO₄</em>
d. 0,034 moles of Li₂Cr₂O₇ are:
0,034 mol Li₂Cr₂O₇× = <em>7,82g of Li₂Cr₂O₇ </em>
e. 0,034 moles of K₂C₂O₄ are:
0,034 mol K₂C₂O₄× = <em>5,65g of K₂C₂O₄</em>
I hope it helps!