ANSWERS TO REVIEW CLASS PROBLEMS, SUN 6/04/06

1) FIRST QUESTION

(1) a-Asked for KNO3(35oC, 50.0g) according to table G it is (35oC, and about 54g) therefore it is unsaturated.

b- Asked about NH3(35oC, 10g) according to the table G, is (350C, about 441g) –unsaturated.

c- Asked for NaCl (35oC, 20g ), according to table G it is , ( 350C, about 38g) – unsaturated

d-Asked for KI (70oC, 40g), according to table G it is (70oC, off scale) –unsaturated)

e-Asked for KNO3 (20oC, 110g), according to table G it is (20oC, about 36g)-supersaturated.

(2) a- M = moles/liter volume

  1. Must first convert KNO3 to moles: MOLES = MASS/MOLAR MASS
  2. = 50 g/101.1034g/mol

round moles to 0.49454 mol

K / 1 * / 39.098 / =39.098
N / 1 * / 14.006 / =14.006
O / 3 * / 15.9994 / = 47.9982

101.1034 g/mol total round to 101.103

  1. MOLARITY = MOLES/LITERS

MOLARITY = 0.49454 mol/ 0.5L ***** NOTE -= 500 ml/ 1000ml/L = 0.5L

MOLARITY = 0.99 molar

(2) b- MOLARITY = MOLES/LITERS

MOLARITY = 3 mol HCl/ 2 L water

MOLARITY = 1.5 M

(2) c – FIRST find the moles carried in 2 L of 1 Molar solution

  1. MOLARITY = MOLES/LITERS

1 molar = MOLES/ 2 L

2 = MOLES

(2) d- A dilution, you change VOLUME but not moles of solute.

MOLARITY(initial) * VOLUME(initial) = MOLARITY(final) * VOLUME(final)

(1 molar) * (1 L) = (X molar) * ( 2 L(

X = 0.5 molar.

(5) parts per million = grams solute/ grams solution x 1,000,000

= 1 gram pesticide / 10,001 grams solution * 1,000,000 = 99.0 ppm

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2) SECOND QUESTION

a-q(heat transfer) = MASS * ∆T oC * SPECIFIC HEAT

= 100g * 40oC * 4.18 J/g*oC

= +16720 J round to 17000 J, divide by 1000 to get 17. kJ

b) q(heat transfer) = MASS * ∆T oC * SPECIFIC HEAT

100J = 100g * ∆T * .239 J/g*oC

.239 oC = ∆T, therefore the final temp is 10.239 oC

c) q(heat transfer) = MASS * ∆T oC * SPECIFIC HEAT

100J = 200g * ∆T * 4.18 J/g*oC

0.119 oC = ∆T therefore the 50 – 0.119 =49.881 oC

Question 5

Q = MASS * ∆Hvap

Q = 1000g * 334 J/g

Q = 223000 J = 334 kJ

Question 6

Element / #moles / Atomic mass / subtotal
K / 2 * / 39.0983 / = 78.1966 g
Cr / 2 * / 51.996 / =101.992 g
O / 7 * / 15.9994 / =111.9958 g
total / 292.1844 g/mol

c) % mass composition = mass part / molar mass (whole) * 100

= K / K2Cr2O7

= 78.1966g / 292.1844 g/mol= 0.26224 * 100 = 26.2% (rounded to three significant figures)

=

d) % mass composition = mass part / molar mass (whole) * 100

= Cr / K2Cr2O7

= 101.992 g / 292.1844 g/mol = 0.34906 * 100 = 34.9% ( round to three significant figures)