What mass of substance would be required to prepare each of the following (answer in each case to four decimal places): (a) \(250.00 \mathrm{~mL}\) of sodium chloride \((0.05 \mathrm{M})\) from \(\mathrm{NaCl} ?\) (b) \(100.00 \mathrm{~mL}\) of potassium iodate \((0.02 \mathrm{M}\) ) from \(\mathrm{KIO}_{3} ?\) (c) \(50.00 \mathrm{~mL}\) of sodium thiosulphate \((0.05 \mathrm{M})\) from \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O} ?\) (d) \(250.00 \mathrm{~mL}\) of copper ions \((0.1 \mathrm{M})\) from \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} ?\) (e) \(100.00 \mathrm{~mL}\) of potassium ions \((0.05 \mathrm{M})\) from \(\mathrm{K}_{2} \mathrm{SO}_{4} ?\)

Step by step solution

01

Solution for (a)

Given that the volume of the solution to prepare is \(250.00 \mathrm{~mL}\) or \(0.250 \mathrm{~L}\) and the molarity is \(0.05 \mathrm{M}\). The molar mass of sodium chloride (NaCl) can be calculated as \(= 22.99 + 35.45 = 58.44 \mathrm{~g/mol}\). By definition of molarity, number of moles \(= Molarity \times Volume = 0.05 \times 0.250 = 0.0125 \mathrm{~moles}\). Thus, the mass of Sodium Chloride required \(= number \,of \,moles \times molar \,mass = 0.0125 \times 58.44 \approx 0.7305 \mathrm{~g}.\)

02

Solution for (b)

Given that the volume of the solution to prepare is \(100.00 \mathrm{~mL}\) or \(0.100 \mathrm{~L}\) and the molarity is \(0.02 \mathrm{M}\). The molar mass of potassium iodate (KIO3 ) can be calculated as \(= 39.10 + 126.90 + 16 \times 3 = 214.00 \mathrm{~g/mol}\). By definition of molarity, number of moles \(= Molarity \times Volume = 0.02 \times 0.100 = 0.002 \mathrm{~moles}\). Thus, the mass of Potassium Iodate required \(= number \,of \,moles \times molar \,mass = 0.002 \times 214.00 \approx 0.4280 \mathrm{~g}.\)

03

Solution for (c)

Given the volume of the solution to prepare is \(50.00 \mathrm{~mL}\) or \(0.050 \mathrm{~L}\) and the molarity is \(0.05 \mathrm{M}\). The molar mass of sodium thiosulphate \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) can be calculated as \(= 2 \times (22.99) + 2 \times (32.065) + 3 \times (15.999) + 5 \times (18.015) = 248.182 g/mol\). By definition of molarity, number of moles \(= Molarity \times Volume = 0.05 \times 0.050 = 0.0025 \mathrm{~moles}\). Thus, the mass of Sodium Thiosulphate required \(= number \,of \,moles \times molar \,mass = 0.0025 \times 248.182 \approx 0.6205 \mathrm{~g}\).

04

Solution for (d)

Given that the volume of the solution to prepare is \(250.00 \mathrm{~mL}\) or \(0.250 \mathrm{~L}\) and the molarity is \(0.1 \mathrm{M}\). The molar mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) can be calculated as \(= 63.55 + 32.065 + 4 \times (15.999) + 5 \times (18.015) = 249.684 \mathrm{~g/mol}\). By definition of molarity, number of moles \(= Molarity \times Volume = 0.1 \times 0.250 = 0.025 \mathrm{~moles}\). Thus, the mass of compound required \(= number \,of \,moles \times molar \,mass = 0.025 \times 249.684 \approx 6.2421 \mathrm{~g}.\)

05

Solution for (e)

Given that the volume of the solution to prepare is \(100.00 \mathrm{~mL}\) or \(0.100 \mathrm{~L}\) and the molarity is \(0.05 \mathrm{~M}\). The molar mass of \(\mathrm{K}_{2}\mathrm{SO}_{4}\) can be calculated as \(= 2 \times (39.10) + 32.065 + 4 \times (15.999) = 174.259 \mathrm{~g/mol}\). By definition of molarity, number of moles \(= Molarity \times Volume = 0.05 \times 0.100 = 0.005 \mathrm{~moles}\). Thus, the mass of Potassium Sulphate required \(= number \,of \,moles \times molar \,mass = 0.005 \times 174.259 \approx 0.8713 \mathrm{~g}.\)

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!