Chapter 7 Solutions And Colloids Worksheet With Answers Page 13
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33Solutions and Colloids 155
SOLUTION PREPARATION (SECTION 7.5)
⎛
⎞⎛
⎞ ⎛
⎞
7.42
a.
1 L
0.250 moles Na SO
142.1 g Na SO
=
2
4
2
4
⎜
⎟⎜
⎟ ⎜
⎟
100 mL
3.55 g Na SO
⎜
⎟⎜
⎟ ⎜
⎟
2
4
1000 mL
1 L solution
1 mole Na SO
⎝
⎠⎝
⎠ ⎝
⎠
2
4
I would weigh 3.55 g Na
SO
and add it to a 100 mL volumetric flask. I would add
2
4
enough water to dissolve the Na
SO
, then add water up to the mark on the volumetric
2
4
flask, cap, and shake to ensure the solution is homogeneous.
⎛
⎞⎛
⎞⎛
⎞
b.
1 L
0.100 moles Zn(NO )
189 g Zn(NO )
=
3 2
3 2
⎜
⎟⎜
⎟⎜
⎟
500 mL
9.45 g Zn(NO )
⎜
⎟⎜
⎟⎜
⎟
3 2
⎝
1000 mL
⎠⎝
1 L solution
⎠⎝
1 mole Zn(NO )
⎠
3 2
I would weigh 9.45 g Zn(NO
)
and add it to a 500 mL volumetric flask. I would add
3
2
enough water to dissolve the Zn(NO
)
, then add water up to the mark on the volumetric
3
2
flask, cap, and shake to ensure the solution is homogeneous.
×
=
×
=
250 g 2.50%(w / w) 250 g 0.0250 6.25 g NaCl
c.
(
)
−
=
=
≈
1 mL
250 g
6.25 g
243.75 g H O
243.75 mL H O 244 mL H O
2
1.00 g H O
2
2
2
I would weigh 6.25 g NaCl and add it to a 400 mL Erlenmeyer flask. I would measure
244 mL of water and add enough of it to the Erlenmeyer flask to dissolve the salt, then I
would add the rest of the water and swirl the flask to ensure that the solution is
homogeneous.
×
=
100 mL 0.55%(w / v) 0.55 g KCl
d.
I would add 0.55 g KCl to a 100 mL volumetric flask. I would add enough water to
dissolve the KCl, then dilute the mixture to the mark on the volumetric flask. I would
then cap the flask and shake it to ensure the solution is homogeneous.
⎛
⎞ ⎛
⎞ ⎛
⎞
7.43
a.
1 L
2.00 moles NaOH
40.0 g NaOH
=
⎜
⎟ ⎜
⎟ ⎜
⎟
500 mL
40.0 g NaOH
⎜
⎟ ⎜
⎟ ⎜
⎟
1000 mL
1 L solution
1 mole NaOH
⎝
⎠ ⎝
⎠ ⎝
⎠
I would weigh 40.0 g NaOH and add it to a 500 mL volumetric flask. I would add
enough water to dissolve the NaOH, then add water up to the mark on the volumetric
flask, cap, and shake to ensure the solution is homogeneous.
⎛
⎞
b.
40.0 mL alcohol
=
⎜
⎟
250 mL solution
100 mL alcohol
⎜
⎟
100 mL solution
⎝
⎠
I would pipette 100 mL of alcohol and add it to a 250 mL volumetric flask. I would add
water up to the mark on the volumetric flask, cap, and shake to ensure the solution is
homogeneous.
⎛
⎞
c.
10.0 g glycerol
=
⎜
⎟
100 mL solution
10.0 g glycerol
⎜
⎟
⎝
100 mL solution
⎠
⎛
⎞⎛
⎞
10.0 g glycerol
1 mL glycerol
=
⎜
⎟⎜
⎟
100 mL solution
7.94 mL glycerol
⎜
⎟⎜
⎟
100 mL solution
1.26 g glycerol
⎝
⎠⎝
⎠
I would either weigh 10.0 g glycerol on a balance or pipette 7.94 mL glycerol into a 100
mL volumetric flask. I would add water up to the mark on the volumetric flask, cap, and
shake to ensure the solution is homogeneous.
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