find the mass of glucose present in 3m of solution
here what is 3m ...m represent
Answers
Explanation:
A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?
Solution
Since the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L:
M
=
mol solute
L solution
=
0.133
mol
355
mL
×
1
L
1000
mL
=
0.375
M
Check Your Learning
A teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL?
Answer:
0.05 M
Example 2
Deriving Moles and Volumes from Molar Concentrations
How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example 1?
Solution
In this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in Example 1, 0.375 M:
M
=
mol solute
L solution
mol solute
=
M
×
L solution
mol solute
=
0.375
mol sugar
L
×
(
10
mL
×
1
L
1000
mL
)
=
0.004
mol sugar
Check Your Learning
What volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?
Answer:
80 mL
Example 3
Calculating Molar Concentrations from the Mass of Solute
Distilled white vinegar (Figure 2) is a solution of acetic acid, CH3CO2H, in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?
A label on a container is shown. The label has a picture of a salad with the words “Distilled White Vinegar,” and, “Reduced with water to 5% acidity,” written above it.
Figure 2. Distilled white vinegar is a solution of acetic acid in water.
Solution
As in previous textbox shaded, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute’s molar mass to obtain the amount of solute in moles:
M
=
mol solute
L solution
=
25.2
g CH
3
CO
2
H
×
1
mol CH
2
CO
2
H
60.052
g CH
2
CO
2
H
0.500
L solution
=
0.839
M
M
mol solute
L solution
=
0.839
M
M
0.839
mol solute
1.00
L solution
Check Your Learning
Calculate the molarity of 6.52 g of CoCl2 (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.
Answer:
0.674 M
Example 4
Determining the Mass of Solute in a Given Volume of Solution
How many grams of NaCl are contained in 0.250 L of a 5.30-M solution?
Solution
The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example 2:
M
=
mol solute
L solution
mol solute
=
M
×
L solution
mol solute
=
5.30
mol NaCl
L
×
0.250
L
=
1.325
mol NaCl
Finally, this molar amount is used to derive the mass of NaCl:
1.325
mol NaCl
×
58.44
g NaCl
mol NaCl
=
77.4
g NaCl
Check Your Learning
How many grams of CaCl2 (110.98 g/mol) are contained in 250.0 mL of a 0.200-M solution of calcium chloride?
Answer:
5.55 g CaCl2
When performing calculations stepwise, as in Example 4, it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example 4, the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.
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