Question

Complete the following table:

	Mass	Mol	Volume
Solute	Solute	Solute	Solution	Molarity
KNO3	23.5 g	_____	135.0 mL	_____
NaHCO3	_____	_____	200.0 mL	0.140 M
C12H22O11	56.38 g	_____	_____	0.150 M

Part A

Complete the row for KNO3.

Part B

Complete the row for NaHCO3.

Part C

Complete the row for C12H22O11.

Answer 1

moles of any substance = weight/molecular weight

and molarity = weightx1000/molecular weight x V (volume)

now moles of KNO3 = weight of KNO3/M.Wt of KNO3

= 23.5/101.103

= 0.232

now molarity of KNO3 = moles of KNO3 x 1000/volume of solution

= 0.232 x1000/135

= 1.72 m

coming to NaHCO3

molarity = weightx1000/molecular weight x V (volume)

0.14 = weight x 1000/84 x200

now weight = 84 x200 x0.14/1000

= 2.35 gm

coming to part-C

weight of C12H22O11. = 56.38 gm

Molecular weight of C12H22O11 = 342

now moles of C12H22O11 = 56.38/342

= 0.1648

molarity = moles x 1000/volume

volume= moles x1000/molarity

= 1098 mL