Question

Be sure to answer all parts.

Determine how many grams of each of the following solutes would be needed to make 4.30 × 102 mL of a 0.100 M solution.

(a) cesium bromide (CsBr):



(b) calcium sulfate (CaSO4):



(c) sodium phosphate (Na3PO4):



(d) lithium dichromate (Li2Cr2O7):



(e) potassium oxalate (K2C2O4):

Answer 1

Answer:

(a) cesium bromide (CsBr): 9.15 grams

(b) calcium sulfate (CaSO4):  5.85 grams

(c) sodium phosphate (Na3PO4): 7.05 grams

(d) lithium dichromate (Li2Cr2O7):  9.88 grams

(e) potassium oxalate (K2C2O4):   7.15 grams

Explanation:

(a) cesium bromide (CsBr):

Molar mass of CsBR = 212.81 g/mol

Number of moles = molarity * volume

Number of moles = 0.100 M *0.43 L

Number of moles = 0.043 moles

Mass of CsBr required = moles * Molar mass

Mass of CsBr required = 0.043 moles * 212.81 g/mol

Mass of CsBr required = 9.15 grams

(b) calcium sulfate (CaSO4):

Molar mass of CaSO4 = 136.14 g/mol

Mass of CaSO4 required = moles * Molar mass

Mass of CaSO4 required = 0.043 moles * 136.14 g/mol

Mass of CaSO4 required = 5.85 grams

(c) sodium phosphate (Na3PO4):

Molar mass of Na3PO4 = 163.94 g/mol

Mass of Na3PO4 required = moles * Molar mass

Mass of Na3PO4 required = 0.043 moles * 163.94 g/mol

Mass of Na3PO4 required = 7.05 grams

(d) lithium dichromate (Li2Cr2O7):

Molar mass of Li2Cr2O7 = 229.87 g/mol

Mass of Li2Cr2O7 required = moles * Molar mass

Mass of Li2Cr2O7 required = 0.043 moles * 229.87 g/mol

Mass of Li2Cr2O7 required = 9.88 grams

(e) potassium oxalate (K2C2O4):

Molar mass of K2C2O4 = 166.22 g/mol

Mass of K2C2O4 required = moles * Molar mass

Mass of K2C2O4 required = 0.043 moles * 166.22 g/mol

Mass of K2C2O4 required = 7.15 grams