Determine how many grams of each of the following solutes would be needed to make 3.40 × 102 ml of a 0.100 m solution. (a) cesium bromide (csbr): g (b) calcium sulfate (caso4): g (c) sodium phosphate (na3po4): g (d) lithium dichromate (li2cr2o7): g (e) potassium oxalate (k2c2o4): g

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× = 7,24g of CsBr -Where 212,81g/mol is molar mass of CsBr-

b. 0,034 moles of CaSO₄ are:

0,034 mol CaSO₄× = 4,63g of CaSO₄

c. 0,034 moles of Na₃PO₄ are:

0,034 mol Na₃PO₄× = 5,57g of  Na₃PO₄

d. 0,034 moles of Li₂Cr₂O₇ are:

0,034 mol Li₂Cr₂O₇× = 7,82g of Li₂Cr₂O₇

e. 0,034 moles of K₂C₂O₄ are:

0,034 mol K₂C₂O₄× = 5,65g of K₂C₂O₄

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