Question

11 Review Constants I Period What is the total number of moles of particles in 1.0 Lof each of the following solutions? Part A You may want to reference (Pages 338 - 347) Section 9.6 while completing this problem. 0.9% (m/v) NaCl (strong electrolyte), used to replace lost body fluids and salts Express the number of moles to one significant figure. IVO AQ R O a ? moles Submit Request Answer Part B 5.0% (m/v) dextrose (C6H206), used to provide fluids and energy for patients Express your answer using two significant figures. O AD O ? moles Submit Request Answer

Answer 1

A)

%(m/v) = mass of NaCl * 100 / volume of solution in mL

0.9 = mass of NaCl * 100 / 1000 mL

mass of NaCl = 9.0 g

Molar mass of NaCl,

MM = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

mass(NaCl)= 9.0 g

use:

number of mol of NaCl,

n = mass of NaCl/molar mass of NaCl

=(9 g)/(58.44 g/mol)

= 0.154 mol

1 NaCl gives 2 particles.

So,

Mol of Particles = 2* 0.154 mol

= 0.308 mol

Answer: 0.3 mol

B)

%(m/v) = mass of glucose * 100 / volume of solution in mL

5.0 = mass of glucose * 100 / 1000 mL

mass of glucose = 50 g

Molar mass of C6H12O6,

MM = 6*MM(C) + 12*MM(H) + 6*MM(O)

= 6*12.01 + 12*1.008 + 6*16.0

= 180.156 g/mol

mass(C6H12O6)= 50 g

use:

number of mol of C6H12O6,

n = mass of C6H12O6/molar mass of C6H12O6

=(50 g)/(1.802*10^2 g/mol)

= 0.2775 mol

Answer: 0.28 mol