13. (a) When 1.00 L sample of water from the surface of the Dead sea is evaporat

Question

13. (a) When 1.00 L sample of water from the surface of the Dead sea is evaporated, 179 g of MgCl2 is recovered. What was the molarity of MgCl2 in the water sample? (b) If the density of the seawater at a depth of 10,000 m is 1.071 g/mL and a 25.0 g sample of water from that depth contains 99.7 mg K^+, what is the molarity of potassium ions in the sample? 14. An aqueous solution called phosphate-buffered saline (PBS) is used in biology research to wash and store living cells. It contains ionic solutes including 10.0 mMNa2HPO4 - 2 H2O. How many grams of this solute would you need to prepare 5.0 L of PBS? 15. The concentration of Pb^2+ in a standard solution is 1.000 mg/mL. What volume of the solution should be used to prepare 500.0 mL of a solution in which the Pb^2+ concentration is 2.50 mg/L?

Explanation / Answer

13. a). Mass of MgCl2 = 179 g

Molar mass of MgCl2 = 95.211 g/mol

moles of MgCl2 = 179 / 95.211

= 1.88

Volume of water = 1.0L

[MgCl2] = 1.88 / 1.0

= 1.88 M

b). Density of water = 1.071 g/mL

Mass of sample = 25.0 g

Volume of sample = 25.0 / 1.071

= 23.34 mL

= 0.02334 L

Mass of K+ ions = 99.7 mg = 0.0997 g

Molar mass of K+ ions = 39.0983 g/mol

Moles of K+ ions = 0.0997 / 39.0983

= 0.002549 moles

[K+] = 0.002549 / 0.02334

= 0.109 M

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