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21 March, 21:41

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) :

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  1. 21 March, 22:00
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    (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
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