Question

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

Answer 1

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×$\frac{212,81g}{1mol}$ = 7,24g of CsBr -Where 212,81g/mol is molar mass of CsBr-

b. 0,034 moles of CaSO₄ are:

0,034 mol CaSO₄×$\frac{136,14g}{1mol}$ = 4,63g of CaSO₄

c. 0,034 moles of Na₃PO₄ are:

0,034 mol Na₃PO₄×$\frac{163,94g}{1mol}$ = 5,57g of  Na₃PO₄

d. 0,034 moles of Li₂Cr₂O₇ are:

0,034 mol Li₂Cr₂O₇×$\frac{229,87g}{1mol}$ = 7,82g of Li₂Cr₂O₇

e. 0,034 moles of K₂C₂O₄ are:

0,034 mol K₂C₂O₄×$\frac{166,2g}{1mol}$ = 5,65g of K₂C₂O₄

I hope it helps!